Compute $\lim_{n\to\infty} n \bigg( \sum_{k=0}^n f(\frac{k}{n}) - n\int_0^1 f(x) \, dx \bigg)$ The task is to show the following limit exists, and then compute it.  Here,
$\,\mathrm{f}:\left[0,1\right] \to \mathbb{R}$ is a continuously differentiable function.
$$
\lim_{n \to \infty}\left\{n\left[%
\sum_{k = 0}^{n}\,\mathrm{f}\left(k \over n\right) - n\int_{0}^{1}\,\mathrm{f}\left(x\right)\,\mathrm{d}x\right]\right\}
$$
It seems so innocent, it should be easy for me at this point$\ldots$ Could anyone provide some direction or intuition on how to approach such problems ?.  Specific hints would be helpful as well.

EDIT:
The comments below discuss why this limit does not exist in general as written.  Perhaps this problem $\left(~\mbox{found on a qualifying exam}~\right)$ had a typo.

We will make some extra assumption on $f(x)$ and shows that under such assumption, the limit diverges unless $f(0)+f(1) = 0$.

Let $f : [0,1] \to \mathbb{R}$ be any $C^2$ function on $[0,1]$, i.e. twice differentiable and the $2^{nd}$ derivative $f''(x)$ is continuous.
For any $0 \le a < b \le 1$, let $h = b - a$ and consider following integral

$$\mathcal{I} \stackrel{def}{=} \int_a^b f''(x) (x-a)(b-x) dx$$
Integrate by parts, it is easy to see

$$\begin{align}
\mathcal{I}
&= \int_a^b (x-a)(b-x) df'(x) =
\left[(x-a)(b-x)f'(x)\right]_a^b - \int_a^b f'(x)(a+b - 2x) dx\\
&= \int_a^b (2x - (a+b)) df(x)
= \left[(2x-(a+b))f(x)\right]_a^b - 2\int_a^b f(x) dx\\
&= h(f(a)+f(b)) - 2\int_a^b f(x)dx
\end{align}
$$
Notice the factor $(x-a)(b-x)$ is non-negative on $[a,b]$ with $\displaystyle\;\int_a^b (x-a)(b-x) dx = \frac{h^3}{6}\;$.
If $M$ and $m$ is the maximum and mimumum of $f''(x)$ over $[a,b]$, we have following bound for $\mathcal{I}$.

$$m\frac{h^3}{6} \le \mathcal{I} \le M\frac{h^3}{6}$$

Since $f''(x)$ is continuous over $[a,b]$, by IVT, there exists an $c \in (a,b)$ such that $$\mathcal{I} = f''(c)\frac{h^3}{6}$$

Now for any fixed $n > 1$, sub divide $[0,1]$ into $n$ subintervals $[a_k,b_k] = \left[\frac{k}{n},\frac{k+1}{n}\right]$ for $0 \le k < n$. Apply result above,
we find there are $c_k \in [a_k,b_k]$ such that

$$\frac{1}{12n^3} f''(c_k) = \frac{1}{2n}\left(f(a_k)+f(b_k)\right) - \int_{a_k}^{b_k} f(x)dx$$
Summing over $k$ give us

$$n\left\{
\sum_{k=0}^{n-1} \frac12\left[
f\left(\frac{k}{n}\right)
+f\left(\frac{k+1}{n}\right)
\right] - n\int_0^1 f(x)dx
\right\} = \frac{1}{12n}\sum_{k=0}^{n-1}f''(c_k)$$

The RHS has the form of a Riemann sum over $f''(x)$. This means as $n \to \infty$,
the limit of LHS exists and equals to
$$\lim_{n\to\infty} n\left\{
\sum_{k=0}^{n-1} \frac12\left[
f\left(\frac{k}{n}\right)
+f\left(\frac{k+1}{n}\right)
\right] - n\int_0^1 f(x)dx
\right\}\\ = \frac{1}{12} \int_0^1 f''(x)dx
= \frac{1}{12} (f'(1)-f'(0))\tag{*1}
$$
Compare this with the sequence in question, we find

$$n\left\{\sum_{k=0}^n f\left(\frac{k}{n}\right) - n\int_0^1 f(x)dx\right\}
= \frac{n}{2}(f(0)+f(1)) + \frac{1}{12}(f'(1)-f'(0)) + o(1)
$$
From this, we find unless $f(0)+f(1) = 0$, the sequence at hand contains a term proportional to $n$ and diverges in general. If $f(0)+f(1) = 0$, the limit do
exists and equals to $\frac{1}{12}(f'(1) - f'(0))$.

As mentioned in comment, this is essentially
Euler Maclaurin formula of approximating the integral over $[0,1]$.

As a side note, I think the condition $f$ is $C^2$ can be relaxed. However, I don't know the exact condition for $(*1)$ to be true. $C^2$ is simply the condition one can prove what one need directly.

Here is a start.

Factor out a $n$ from inside and rearrange things and you get the following

$\lim_{n \rightarrow \infty} \frac{ \frac{1}{n} \sum_{i=1}^n f(\frac{k}{n}) - \int_0^1 f(x) dx}{\frac{1}{n^2}} = \frac{0}{0}$

The top zero is because you have a Riemann approximation (the summation) which approaches exactly your integral.  So from here we could do l'hospital's rule (you just have to be careful with doing derivatives of summations that depend on the variable)


What is the period of the $f(x)=\sin x +\sin3x$? What is the period of the $f(x)=\sin x +\sin3x$?

$f(x)=\sin x+\sin 3x=2\frac{3x+x}{2}\cos\frac{x-3x}{2}=2\sin2x\cos x=4\sin x\cos^2x\\f(x+T)=4\sin(x+T)\cos^2(x+T)=4\sin x\cos^2 x$

how can I deduct this I have no idea

In general, if $T$ is the period of a function $f(x)$ then the period of the function $f(ax)$ is $\frac{T}{a}$.

Suppose two periodic functions $f_1(x)$ and  $f_2(x)$ have periods $T_1$ and $T_2$. Then the period of the function $g(x)=f_1(x)\pm f_2(x)$ is LCM (least common multiple) of $T_1$ and $T_2$ (although, this certainly isn't true for all periodic functions, as explained inside this answer.)

In your question the periods of $\sin x$ and $\sin 3x$ are calculated as $\frac{2\pi}{1}=2\pi$ and $\frac{2\pi}{3}$ respectively.

So the period of the function $f(x)=\sin x+\sin3x$ is the $\text{LCM}(2\pi,\frac{2\pi}{3})=2\pi$.

The period of $sin(x)$ is $2\pi$ and the period of $sin(3x)$ is $\displaystyle \frac{2\pi}{3}$. The period is thus $2\pi$.


What are the continuous functions $ x f(y)+y f(x)=(x+y) f(x) f(y) ? $ question -

What are the continuous functions on $\mathbb{R}$ which are solutions of the equation
$$
x f(y)+y f(x)=(x+y) f(x) f(y) ?
$$

my try -

by putting $y=x$ i get $f(x)=0$ or $1$ for all $x$ not equal to $0$...
now my answer is same as mention in book but i think it is incorrect because they are not valid for all $x$ ???

can someone tell how to fix this hole with the help of continuity ...
i know this is simple question but i want to clear my doubt...

thankyou

Wit $x=y$, we arrive at the necessary condition
$$2xf(x)=2xf(x)^2 $$
and hence $$\tag1f(x)\in\{0,1\}\quad\text{for }x\ne0. $$
The only continuous functions with this property are the constant functions
$$f(x)=0$$
and
$$f(x)=1. $$
Both are directly verified to indeed solve the original functional equation.

What if we drop continuity?

We still have $(1)$, but note that $f$ may jump discontinuously between $0$ and $1$.
Suppose $f(x_0)=0$ for at least one $x_0\ne 0$. Then with $y=x_0$, we get
$x_0f(x)=0 $
and hence
$$ f(x)=0\quad\text{for all } x$$
as one solution (again).

So assume $f(x)=1$ for all $x\ne 0$.
Then nothing can be said about $f(0)$, i.e., for any $c$,
$$f(x)=\begin{cases}c&x=0\\1&x\ne 0\end{cases} $$
is a solution, as one readily verifies.

Suppose $f(x) =0$ for all $x \neq 0.$ Since $f$ is continuous, so $f(0)=\lim_{x \to 0} f(x)=0$. Hence $f(x)=0$ for all $x \in \mathbb{R}.$

Similarly, the other case follows.


Expected value of Dirac measure: $E[\delta_{\tau}(t)]$ Thx for the efforts! I tried to clean up the question and remove the ambiguous notations. For the dirac measure, I use the one found in the Wikipedia article.

I try to formally understand a statement found across literature (as @Did pointed out, assuming independence):

$E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] = \int_{0}^{T}E[D(t)]dP_{\tau}(t)$.

To derive it, the authors do the following:

$E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] = E[\int_{-\infty}^{\infty} \mathbb{I}_{[0,T]}(t) D( t) d\delta_{\tau}(t)] = E[\int_{0}^{T} D( t) d\delta_{\tau}(t)] = E[\int_{0}^{T} D( t) \delta_{\tau}(t)dt]$.

And then, what I don't get, they do the following (which might be a notational mess):

$\int_{0}^{T} E[D(t)] E[d\delta_{\tau}(t)] = \int_{0}^{T} E[D(t)] P(\tau \in [t,dt]) $

There are two questions

How does the distribution of $\tau$ enter the picture, i.e. what is the relation of $\delta_{\tau}$ to $P_{\tau}$, to get the proposition (1st equation) from the second equation? (MickG's explanations).
What is going on with the third equation? Which is hopefully resolved after answering the first question:-) Maybe, the authors mean something like $E[\delta_{\tau}(t)] = E[\mathbb{I}_{[t,dt]}(\tau)]=P(\tau \in [t,dt])$...

Thx!

On the RHS I get $\int_0^\infty E[D(t)]\mathbb1_{0<t<T}d\mu(t)=\int_0^TE[D(t)]d\mu(t)$. On the LHS, I get $E[E[D(t)]\mathbb1_{0<t<T}]$. So:

$$\int_0^TE[D(t)]d\mu(t)=E[E[D(t)]\mathbb1_{0<t<T}].$$

Seems we have one $E$ too many on the right… If I consider $D(t)\mathbb1_{[0,T]}(t)$ instead, I get:

$$E[D(t)\mathbb1_{[0,T]}(t)]=\int_0^\infty D(t)\mathbb1_{[0,T]}(t)d\mu(t)=\int_0^TD(t)d\mu(t).$$

Now there is a missing $E$. So the question I ask is: is the identity stated in your reference actually correct? Also, that integral of $E[D(t)]$ leaves me wondering whether $E[D(t)]$ actually depends on $t$. After all, it is an expected value, so why should it depend on anything? With that, I would have that the first equality has LHS equal to $P([0,T])E[D(t)]$.

Let us now look at how your authors derive their identity:

\begin{align*}
E[\mathbb{I}_{[0,T]}(\tau)D(\tau)] ={}& E\left[\int_{-\infty}^{\infty} \mathbb{I}_{[0,T]}(t) D( t) \delta(t-\tau)dt\right] = E\left[\int_{0}^{T} D( t) \delta(t-\tau)dt\right] ={} \\
{}={}& \int_{0}^{T} E[D(t)] E[\delta(t-\tau)]dt = \int_{0}^{T} E[D(t)] P(\tau \in [t,dt]).
\end{align*}

I restored the "$\delta$ function" since this is just a formal trick to convince readers of the equality. Let us try to see this formal derivation step by step:

Introduce the $\delta$ function to express evaluation as an integral;
Collapse the indicator into the integration bounds;
Bring the $E$ in (it is, at the end of the day, an integral, so you can surely swap it with $\int_0^T$, right?), then assume some sort of independence between $D$ and $\delta$ to turn $E$ of a product to a product of $E$; now this is very sloppy;
A very big problem; It seems they assume $E[\delta(t-\tau)]=P(\tau\in[t,dt])$; why is that? And what does that probability even mean? After all, infinitesimal are out of standard analysis, so what is $dt$? They are probably thinking of having $\tau$ very close to $t$; or, they may be thinking of approximating the $\delta$ function with some more treatable function; if $\delta$ is 0 everywhere except $\delta(0)=\infty$, we can see $\delta(x)=\lim_{n\to\infty}n\mathbb1_{[-\frac1n,\frac1n]}(x)$; so $\delta(t-\tau)=\lim_nn\mathbb1_{[-\frac1n,\frac1n]}(t-\tau)$; assuming limit and integral swap, $E[\delta(t-\tau)]=\int\delta(t-\tau)d\mu(\tau)=\lim_n\int n\mathbb1_{[-\frac1n,\frac1n]}(t-\tau)d\mu(\tau)=\lim_nnP(\tau\in[t-\frac1n,t+\frac1n]$; now $\tau$ has a density, and we know the density is the derivative of the cumulative distribution function, so since $P(\tau\in[t-\frac1n,t+\frac1n])=F_P(t+\frac1n)=F_P(t-\frac1n)$, where $F_P$ is the CDF, we get precisely $f(t)$ as that expected value of the $\delta$.

To better (but always formally) treat the third step, we may assume Fubini-Tonelli works fine and try this:

$$E\left[\int_0^TD(t)\delta(t-\tau)dt\right]=\int_0^\infty\int_0^TD(t)\delta(t-\tau)dtd\mu(\tau)=\int_0^T\int_0^\infty\delta(t-\tau)d\mu(\tau)D(t)dt,$$

then we have that inner integral, and, if $d\mu(\tau)=f(\tau)d\tau$, we know that is $f(t)\mathbb1_{[0,\infty)}(t)$, and we are there.

I hope this answers your question. Though that identity in your reference seems incorrect, as I pointed out at the start considering two different functions.


Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$ How can I solve this question with out find a,b,c,d

$$a+b+c+d=2$$

$$a^2+b^2+c^2+d^2=30$$

$$a^3+b^3+c^3+d^3=44$$

$$a^4+b^4+c^4+d^4=354$$

so :$$\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}=?$$

If the qusetion impossible to solve withot find a,b,c,d then is there any simple way to find a,b,c,d

Is there any help?

One way to approach such questions is to view $a, b, c, d$ as roots of a single quartic equation $x^4-s_1x^3+s_2x^2-s_3x+s_4=0$, when we have Vieta's relations
$$s_1=a+b+c+d$$$$s_2=ab+ac+ad+bc+bd+cd=(a+b)(c+d)+ab+cd$$$$s_3=abc+abd+acd+bcd$$$$s_4=abcd$$

We then let $p_k=a^k+b^k+c^k+d^k$ with: $$p_0=4$$$$p_1-s_1=0$$$$p_2-p_1s_1+2s_2=0$$$$p_3-p_2s_1+p_1s_2-3s_3=0$$

With the given values of $p_1, p_2, p_3$ these determine $s_1, s_2, s_3$.

For higher powers $(k\ge4)$ we have $$a^k-s_1a^{k-1}+s_2a^{k-2}-s_3a^{k-3}+s_4a^{k-4}=0$$ using the quartic equation. Adding together the corresponding equations for $a , b, c, d$ we obtain the recurrence for $p_k$$$p_k-s_1p_{k-1}+s_2p_{k-2}-s_3p_{k-3}+s_4p_{k-4}=0$$

Since we know $p_4$ we can use the recurrence to find $s_4$, and this is then sufficient to compute $p_5, p_6$ from the recurrence.

If this looks complicated, this is misleading. The relations between the elementary symmetric functions $s_r$ and the sums of powers $p_r$ are useful to know, and render this kind of problem essentially mechanical. You can see that you can solve the problem without solving explicitly for $a, b, c, d$.

Following this scheme we find successively $s_1=2, s_2=-13, s_3=-14, s_4=24$. Then the recurrence gives:$$p_5-2p_4-13p_3+14p_2+24p_1=0$$ and$$p_6-2p_5-13p_4+14p_3+24p_2=0$$

And this gives $p_5=812, p_6=4890$

If you inspect the system you can see that the solutions for a,b,c,d are all going to have to be small integers. This constrains the system heavily.
 I managed to solve the system guessing some small numbers, testing them and tweaking minus signs to get the desired result.
After we have found a,b,c,d the fraction becomes trivial.

Any analytical method to solving this system would have to accommodate high order solutions and would inevitably be very complicated. I would like to see analytical solution but doubt any could be as efficient as trial and error in this case.


Inverting a Linear Fractional Transformation I have a particular Linear Fractional Transformation from $\mathbb{C}$ to $\mathbb{C}$, I am using to solve Laplace's equation, and I was hoping to find the inverse of this transformation.  Is there a algorithm that can be used to find the inverse of a Linear Fractional Transformation?  Are there any useful algorithms for special cases?

The inverse of the linear fractional transformation
$$z\to \frac{a z + b}{c z + d}$$ is the linear fractional transformation
$$z\to \frac{e z + f}{g z + h}$$ where the matrix
$$\begin{pmatrix}e & f\\ g & h\end{pmatrix}$$ is the inverse of the matrix
$$\begin{pmatrix}a & b\\ c & d\end{pmatrix}.$$

Answer is here:

https://ieeexplore.ieee.org/document/7106488/

There is a simple expression to compute the inverse LFT...


Determine where h(x) := x $\lvert x \rvert$ is differentiable from R to R. Determine where h(x) := x $\lvert x \rvert$ is differentiable from R to R.

Not totally sure how to start this.

Much appreciation,

Jesse

If at all possible. The first thing to do is graph the function and see if it looks differentiable.
 Let $f(x) = x |x|$

Case 1. If $x > 0$, then $f(x) = x^2$, which is differentiable.
So $f(x)$ is differentiable on $(0,\infty)$

Case 2. If $x < 0$, then $f(x) = -x^2$, which is differentiable. So $f(x)$ is differentiable on $(-\infty, 0)$

Case 3. If $x = 0$, then the derivative as $x\rightarrow 0^+=0$ and the derivative as $x\rightarrow 0^-=0$. So $f'(0) = 0$.

Hence f is differentiable on all of $\mathbb R$.

The only possible place where $h(x)$ may fail to be differentiable is $x=0$.  So consider
$$lim_{x \to 0} \frac{x|x|}{x}.$$
Does this limit exist? If so, $h(x)$ is differentiable everywhere.


Show $11^{11}+12^{12}+13^{13} =10k$ without direct calculation Prove that $11^{11}+12^{12}+13^{13}$ is divisible by $10$.
Obviously you could just put that in to a calculator and see the results, but I was wondering about some of the other approaches to this? I have not studied modulus', so if you could explain it without them, it would be better for me. Thanks!

$13^4$ ends with $1$ so does $13^{12}$ and thus $13^{13}$ ends with $3$

$12^4$ ends with $6$ so $12^{12}$ also ends with $6$

$11^{11}$ ends with $1$ so your statement is valid.

Assuming you have knowledge of modular arithmetic.

$11\equiv 1\mod 10\implies 11^{11}\equiv 1\mod 10$.

$12\equiv 2\mod 10\implies 12^6\equiv 2^6 \mod10\equiv 4\mod 10$ $\implies 12^{12}\equiv 4^2\mod 10\equiv 6\mod 10$.

$13\equiv 3 \mod 10\implies 13^4\equiv 3^4 \mod 10\equiv 1 \mod 10\implies 13^{13}\equiv 3\mod 10.$

So $11^{11}+12^{12}+13^{13}\equiv 1+6+3 \mod 10\equiv 0 \mod 10.$

Hence $11^{11}+12^{12}+13^{13}$ is divisible by 10.


Flux through cone and hyperboloid I want to find the flux of the vector field
$$F(x,y,z)=(x+z,2x+2y,3y+3z)$$
through the body defined by

$$x^2/16+y^2/4-z^2/3\le 1 \quad\mbox{with} \quad 0\le z \lt 3$$
$$x^2/16+y^2/4-(z-7)^2/4\le 0 \quad \mbox{with}\quad 3\le z \lt 7.$$

I tried to apply the divergence theorem to calculate the flux throw the cone but the integral becomes too difficult because of the limits.

As you already found for this vector field, $ \ \nabla  \cdot  \mathbf{F} \ = \ 6 \ $ , and it is entirely understandable that one would wish to find the flux through this compound surface by applying the Divergence Theorem.

The graph above (exposing the interior of the enclosed volume) presents the situation:  a hyperboloid of one sheet, with its "throat" aligned on the $ z-$ axis and having an elliptical cross-section, meets the lower nappe of a cone of elliptical cross-section, also with its symmetry axis on the $ \ z-$ axis, on the plane $ \ z = 3 \ $ ; there, the two surfaces meet in the same ellipse.

All of the "horizontal" cross-sections of these surfaces are similar ellipses, as we will demonstrate shortly.  Because of the axial alignment of the surfaces, we can reduce the volume calculation to integration in the single variable $ \ z \ $ .  It will be helpful to write the equations of the surfaces as

$$ \frac{x^2}{16} \ + \ \frac{y^2}{4} \ = \ 1 \ + \frac{z^2}{3} \ = \ \sigma_h^2(z) \ \ \ \text{and} $$

$$ \frac{x^2}{16} \ + \ \frac{y^2}{4} \ = \ \frac{(z-7)^2}{4} \ = \ \sigma_c^2(z)\ \ , $$

where we will call the expressions on the right-hand sides of the equations "scaling factors" for the hyperboloid and the cone.  If we then write the surface equations as

$$ \frac{x^2}{16\sigma_h^2} \ + \ \frac{y^2}{4\sigma_h^2} \ = \ 1  \ \ \text{and} \ \  \frac{x^2}{16\sigma_c^2} \ + \ \frac{y^2}{4\sigma_c^2} \ = \ 1 , $$

we see that the cross-sections are ellipses centered on the $ \ z-$ axis, with "major axes" parallel to the $ \ x-$ axis of lengths $ \ 4 \ \sigma_h \ $ and $ \ 4 \ \sigma_c \ $ , and "minor axes" parallel to the $ \ y-$ axis of lengths $ \ 2 \ \sigma_h \ $ and $ \ 2 \ \sigma_c \ $ , respectively.  The infinitesimal volume "slices" along the $ \ z-$ axis are thus $ \ 8 \ \pi \ \sigma_h^2 \ $ and
$ \ 8 \ \pi \ \sigma_c^2 \ $ . [We can easily check that the ellipses "match up" at $ \ z = 3 \ $ , where $ \ \sigma_h^2 \ = \ \sigma_c^2 \ = \ 4 \ $ . ]

We can now write the integral for the volume enclosed by this compound surface as

$$ 8  \pi \ \ \left[ \ \int_0^3  \sigma_h^2(z) \ \ dz \ \ + \ \ \int_3^7  \sigma_c^2(z) \ \ dz \ \right] $$

$$ = \ \ 8  \pi \ \ \left[ \ \int_0^3  \left( 1 \ + \ \frac{z^2}{3}  \right) \ \ dz \ \ + \ \ \int_3^7  \frac{(z-7)^2}{4} \ \ dz \ \right] $$

$$ = \ \ 8  \pi \ \ \left[ \   \left( z \ +  \frac{z^3}{9}  \right) \ \vert_0^3  \ \ + \ \  \left( \frac{(z-7)^3}{12} \right)  \ \vert_3^7  \ \right] $$

$$ = \ \ 8  \pi \ \ \left[ \   \left( 3 \ +  \frac{3^3}{9}  \right)   \ \ - \ \  \left( \frac{(3-7)^3}{12} \right)  \ \right] \ = \ 8  \pi \  ( \ 3 \ + \ 3      \ + \  \frac{64}{12} \ ) \ = \ \frac{272 \pi}{3} \ \ . $$

Hence, by the Divergence Theorem,

$$ \iint_S \ \mathbf{F} \cdot {n} \ \ dS \ \ = \ \ 6 \ \iiint_V \ \nabla \cdot \mathbf{F} \ \ dV \  =  \ 6 \ \cdot \ \frac{272 \pi}{3} \ = \ 544 \pi \ \ . $$

Hint: After converting it into a triple integral via divergence theorem, apply a "polar" change of coordinates of the form $x = 4r \cos{\theta}, y = 2 r \sin{\theta}$. You want to use this change of coordinates because the cross-sections of the given surface are ellipses. Do not forget that you'll also have to calculate the Jacobian for this change of coordinates.


What is the Jacobian matrix of the following. $$
\frac{dx}{dt} = \left(1−x−\frac y2\right)x
\\
\frac{dy}{dt} = \left(1- \frac x3−y\right)y
$$
I am unsure about his to take the partial derivatives of these differential equations

$$
\begin{array}{cc}
  \frac{\partial}{\partial x} (dx/dt) = 1-2x-y/2, &
  \frac{\partial}{\partial x} (dy/dt) = -y/3 \\
  \frac{\partial}{\partial y} (dx/dt) = -x/2, &
  \frac{\partial}{\partial y} (dy/dt) = 1 - x/3 -2y. \\
\end{array}
$$


How to explain to a 14-year-old that $\sqrt{(-3)^2}$ isn't $-3$? I had this problem yesterday. I tried to explain to the kid this: $$\sqrt{(-3)^2} = 3,$$ and he immediately said: "My teacher told us that we can cancel the square with the square root, so it's $$\sqrt{(-3)^2} = -3."$$

He has a lot of problems with maths, and I don't know how I can explain to him this as easily as possible. He still thinks that I lied him.

Thank you.

After having given him the square root function definition, it may be helpful to show the following:
$$
\sqrt{(-3)^2}=\sqrt{9}=\sqrt{3^2}=3\neq -3.
$$
Although well-definedness may be a bit of a heavy topic for a 14-year old, it should be very easy to explain why you cannot have ambiguous definitions in mathematics.

I think perhaps the boy is right and You've not told him the whole truth. $\sqrt{(-3)^2}$ does actually equal -3 but it also equals +3. In the same way as $\sqrt{9}$ = 3 or -3.

I wonder if he has a grasp of negative numbers yet? If not then I think that would be the place to start. If he does then start by showing him that $3^2 = 9$ and $(-3)^2 = 9$, then you can tell him that $\sqrt{9}$ takes us back from the square to either of the original numbers.


Is a line parallel with itself? Simple Question, but I'm finding a lot of dispute on the "lesser" internet.

Basically, given a line, is it parallel with itself?

It's sometimes hard for people learning mathematics, who naturally feel that mathematics is an objective discipline, to hear that many things are actually a matter of convention. This is one example of that, though.  A quick scan of google books will show you that different authors use different definitions for "parallel", and that some of these definitions allow a line to be parallel to itself, while others don't.

There is a second issue in mathematical English that's relevant here. In advanced mathematics, when we say "two objects", we leave open the possibility that the two objects are actually equal. So for example, when I say "the sum of two even numbers is even" I am not requiring the numbers to be distinct.  If I want the objects to be different I have to say "two distinct objects".

However, it appears to me that some of the geometry books I see on google books don't follow this convention. This isn't surprising to me, because

Euclidean geometry has been relegated, to some extent, as a course for pre-service teachers rather than pre-service research mathematicians, and so the audience is not as mathematically advanced.
Euclidean geometry is likely to continue to carry along more traditional phrasing (e.g. from thousands of years in the past) which treated identity as a special relationship. We also have this in English: the reason we traditionally say "I am he" instead of "I am him" is because "to be" was viewed as special and not as a transitive verb.

In any event, the variety of definitions and language conventions underscores the fact that you have to take the definitions of a book in the context of that book, and that you have to make sure that you understand the implicit language conventions (or lack thereof) used by the author.

Two straight lines bearing a constant distance is always parallel... according to that if two lines overlap each other then they have the constant distance zero..so they can be parallel


Is the set of all mathematical truths countable or uncountable? Is the set of all theorems countable or uncountable?

Maybe its a stupid question. I just wanted to know. I am led to think that since, we use a finite set of symbols and English letters, the set of theorems is countable.

EDIT: The set of theorems is not just a set of linguistic combinations, but a set of mathematical truths.

A theorem is, by definition, a finite string of symbols that can be derived by some specified proof system. Because there are countably many finite strings of symbols, there are at most countably many theorems in any given theory.

Speaking of "a set of mathematical truths", where the elements of that set is supposed to be something different from symbolic representations, is not well defined. What kind of object is "a mathematical truth" to you such that you can put them into a set and count them? There is no definition of such a thing in general use, except for the formulas of symbolic logic.

In model theory one can speak of theories where the set of possible symbols are uncountable, such as a symbol for each real number. Such a theory, of course, has uncountably many theorems of the form $c=c$. However, these theories are generally considered artificial objects of study. Studying them can be useful as a stepping stone in proving things about ordinary countable theories, but their formulas are not considered to directly represent "mathematical truth".

The set of all theorems is countable. The set of all mathematical truths isn't. You even could see this like a very basic proof of Gödel incompleteness: It's impossible to prove all mathematical proofs since there are less theorems than truths (which are different things). You can find this question and more interesting stuff on this topic in Geoffrey Hunter's book: Metalogic: An Introduction to the Metatheory of Standard First Order Logic.


How to solve limits? The above limit was solved by making a seemingly arbitrary substitution. The previous limit was solved by making a linear substitution $y=mx$. Which again seemed a bit out of the blue. For another question, my book somehow came to the conclusion that the limit exists and that we should be trying to prove this (again, no explanation was given as to why they were trying to prove the limit existed this time). They then somehow came to the conclusion that a polar coordinate substitution might help along with the Squeeze theorem.
When given a limit, my book keeps using all these different methods from all these different areas of math- most of which are very non-obvious.
So my question(s) boils down to:

a) When given a limit, what's a good way to get "a hunch" if the limit exists or not? I don't want to waste 15 minutes trying to prove a limit that doesn't exist.
b) If I believe the limit exists, what's a good way to approach the problem and generate ideas on how to prove it?
c) If I believe the limit doesn't exist, what's a good way to approach the problem and generate ideas on how to prove it?

These questions obviously don't have deterministic answers that always work, I'm just looking for something to get past the initial "What the hell do I do?!?!". Most of the math I've done so far has been pretty mechanical (keep trying methods from your toolbox until one finally works), so these limits are pretty intimating.

The main geometric idea behind multivariable limits is that they can't depend on the path we take to get there. So to consider $\lim_{(x,y) \to (0,0)} f(x,y)$ we really are considering $\lim_{t \to 0^+} f(x(t),y(t))$ where $x(t),y(t)$ are arbitrary continuous functions (say on $[0,1]$) with $x(0)=y(0)=0$. It's actually a little better than that, because we are permitted to pick any parametrization of the path that we want. So if our path, for example, goes monotonically from $x=0$ to $x=1$ then we can safely take $x(t)=t$, but then $y(t)$ is determined by this choice.

A convenient way to do this is to switch to polar coordinates, because then $(x(t),y(t)) \to (0,0)$ is just $r(t) \to 0$. So we can choose the parametrization $r(t)=t$ and then have $\theta(t)$ be arbitrary. If the limit doesn't depend on the function $\theta(t)$ then it exists, otherwise it doesn't. The linear function test done in the OP amounts to picking $\theta(t)$ to be different constants.

Here are some hopefully useful comments:

If the limit is of the form "two variable polynomial/two variable polynomial" and $(x,y)\to(0,0)$, then you can be quite sure if the limit exists or not. See what is the "degree" of the numerator and the denominator in the following sense:

Degree of $x^3y^2$ is $5$.
Degree of $x-y$ is $1$.
Degree of $x^4+y^2$ is $4$.

If the degree of the numerator is strictly larger than the one of the denominator, the limit exists. If it is the same or smaller, then it doesn't.

If you suspect the limit exists, perhaps the best try is the squeeze theorem. For this you probably need some simple inequalities... Honestly, I cannot remember ever using anything other than the squeeze theorem.

For proving the limit doesn't exist, I like to use sequences in $\mathbb{R}^2$. For example, to show that the limit $\lim_{(x,y)\to(0,0)}\frac{3xy^2}{x^2+y^4}$ doesn't exist, we need two sequences in $\mathbb{R}^2$ that approach $(0,0)$ and that give different results.
Here, I would take $(\frac{1}{n^2},\frac{1}{n})$ and $(\frac{1}{n^2},\frac{2}{n})$. The first one gives $\frac{3}{2}$, and the second one $\frac{12}{17}$. Why did I choose these sequences? I think the best way to choose is to gain some symmetry. Because of $x^2$ and $y^4$, we choose $(\frac{1}{n^2},\frac{1}{n})$, so we get $\frac{2}{n^4}$ in the denominator. Once you choose the exponents for $n$ that give a finite number other than $0$, it is usually easy to find another sequence that gives a different result, thus proving that the limit doesn't exist.

I am sure this is not a complete answer and there is probably more useful stuff I know, but it is hard to talk generally about limits of multivariable functions. If you have some concrete problems, I could look at them and tell you my thoughts on them - not just the solution.


Combinatorics Statistics Question The problem I am working on is:

An academic department with five faculty members—Anderson, Box, Cox, Cramer, and Fisher—must select two of its members to serve on a personnel review committee.
Because the work will be time-consuming, no one is anx-ious to serve, so it is decided that the representative will be selected by putting the names on identical pieces of paper
and then randomly selecting two.

a.What is the probability that both Anderson and Box will
be selected? [Hint:List the equally likely outcomes.]

b.What is the probability that at least one of the two members whose name begins with C is selected?

c. If the five faculty members have taught for 3, 6, 7, 10,
and 14 years, respectively, at the university, what is the
probability that the two chosen representatives have a
total of at least 15 years’ teaching experience there?

For a), I figured that since probability of Anderson being chosen is $1/5$ and Box being chosen is $1/5$ the answer would simply be $2/5$. It isn't, though. It is $0.1$ How did they get that answer? I might need help with parts b) and c) as well.

(a) You will have to consider the 10 different outcomes. Each combination of $"A-B", "B-Cox", "B-Cramer", ...$ is equally likely to be drawn. There are
 $${5 \choose 2} = 10 $$

10 ways to choose 2 people from a group of 5. Since we are only considering 1 possibility, the answer will be $\frac 1 {10} = 0.1$.

For (b), there are a few possibilities where at least 1 "C" member gets selected. Consider $$"Cox-A", "Cox-B", "Cox-Cramer", "Cox-F"$$ and $$"Cramer-A", "Cramer-B", "Cramer-Cox", "Cramer-F"$$ Notice that $"Cox-Cramer"$ appears twice, so one needs to be eliminated. Turns out there are 7 different possibilities out of 10...so we have $0.7$.

Alternatively, you can consider the different ways no "C" member gets selected. Turns out there are 3 ways  $$"A-B", "A-F", "B-F"$$ resulting in $0.3$ chance of the compliment case happening. Taking  $1 - 0.3$, you will still get $0.7$ as the answer.

For (c), I'd personally recommend taking the "compliment" of events. ie, how many combinations have LESS than 15 years? Turns out there are 4 combinations - $(3,6), (3,7), (3,10) , (6,7)$. You can use the answer from (a) and evaluate this to become $1 - \frac 4 {10} = 0.6.$

most simple way to understand this problem.
(i just did this problem just now just now) lol

A, B, Co, Cr, and F exist.
pick 1 of 5 at random. (1/5)
pick another at random, but this time around there are only 4 choices. so (1/4)
multiply the two values. (1/20)

BUT! that's considering that A is picked at first try then B on the second.
you also have to consider the possibility that B would be picked first and etc.

so you add (1/20) + (1/20) = (2/20) = (1/10) = 0.1

pce


Does this property of limits hold when we compose functions? Suppose we have two functions $f,g:\mathbb{R}\to{\mathbb{R}}$ such that:
$$\lim_{x\to{}\alpha}{f(x)}=L \text{ and }\lim_{y\to{\beta}}{g(y)}=\alpha.$$
Is then true that:
$$\lim_{y\to{}\beta}{f(g(y))=L}?$$
My intuition tells me yes. Using an $\epsilon-\delta$ approach seems very long and tedious. Is that the case? Or is there a nice way of showing this to be true. Or, is there a nice counterexample?

Suppose $\alpha\notin g(\mathbb R\setminus\{\beta\})$.
Let $\epsilon>0$ be arbitrary. Then there exists a $\delta>0$ such that $0<|x-\alpha|<\delta\Rightarrow|f(x)-L|<\epsilon$.
Furthermore, there exists a $\eta>0$ such that $0<|y-\beta|<\eta\Rightarrow|g(y)-\alpha|<\delta$. Since $g(y)\ne\alpha$ by our hypothesis, we have $0<|g(y)-\alpha|<\delta$. Then, by my statement above, this implies that $|f(g(y))-L|<\epsilon$, which shows that $\lim_{y\to\beta}f(g(y))=L$.
P.S.: You can get away with assuming that for a small enough $\epsilon>0$ there are no $x\in\mathbb R\setminus\{\beta\}$ with $|x-\beta|<\epsilon$ such that $\alpha=g(x)$.
When this is not the case, there is a sequence of points $y_n\in\mathbb R\setminus\{\beta\}$ with $y_n\to\beta \ (n\to\infty)$ such that $g(y_n)=\alpha$. Then, $\lim_{n\to\infty}f(g(y_n))=f(\alpha)$, so that $\lim_{y\to\beta}f(g(y))=f(\alpha)$ if this limit exists.


Find all values of $n$ such that $\varphi(n) = n/6$. Using the product formula (the formula with the prime factors of $n$), I got
$$1=6\frac{(P_1-1)}{P_1}\frac{(P_2-1)}{P_2}\cdots\frac{(P_k-1)}{P_k}\,.$$

i don't think the proof is enough, as it stands. There is a small ingredient. We are taking the product, with one or more primes, of $\frac{p-1}{p}.$

There are two types of potential problems. Some cancellation is possible, putting the product in lowest terms may be a bit surprising. Second, we can get the product arbitrarily close to $\frac{1}{6}$ byt choosing many large primes.

However: the collection of chosen primes is finite, and one of them is largest, call that $q.$ For any of the other, smaller primes, since $p<q,$ we also have $p-1 < q.$ That means that the numerator of the product fraction is not divisible by $q.$ When we put the fraction in lowest terms, the denominator is still divisible by $q.$

For your problem, this means that the largest prime involved, $q,$ must be a factor of $6.$ The possibilities for the product are just:
$$ \{ \}, \frac{1}{1}   $$
$$ \{ 2\}, \frac{1}{2}   $$
$$ \{ 3\}, \frac{2}{3}   $$
$$ \{ 2,3\}, \frac{1}{3}   $$

Hints:

We know that if $\;p_1,...,p_k\;$ are the different primes diving $\;n\;$ , then

$$\varphi(n)=n\prod_{j=1}^k\left(1-\frac1{p_j}\right)$$

Well, take it now...And I couldn't completely understand what you wrote, but I think you're on the right track: there is no solution.


"Real"-life applications of algebraic geometry Before you tell me that this question has been asked, give me a bit of your time please to read this question because it is not as simple as it sounds.

I did my undergraduate degree in mathematics, taking a pretty heavy course load in theoretical math and doing really well in it. I decided not to proceed with math and am continuing on to a professional degree.

However, every once in a while I have doubts about my decision because it was made on more than one basis, i.e. passion for a given academic subject. One of the reasons was that, seeing my professors, it seemed that mathematicians were very much living in a world of their own and every time I asked for an application of what I was studying, e.g. Galois theory, I got two sorts of answers:

An application of the concept in another area of mathematics...which is not what I was looking for.
A trivial application where a physical/computational/etc system is "modelled" with the concept, e.g. something is a "group", but the recognition that it is was completely useless since the application did not produce a result that would have been otherwise unknown.

My question is: If I changed my mind, applied to do a graduate degree in mathematics and decided to work in a field outside of academia, would I have useful applications of what I studied (and not just a tiny fraction of what I studied, e.g. ODEs) in "real" life"?

I'm very much interested in algebraic geometry (and I am being honest when I say that it is one of the rare things that makes me truly giddy thinking about it). I think you would really answer my question if you could give me an example of a real life problem (not in excessive detail) that was solved thanks to techniques of algebraic geometry. I don't think I have the knowledge of AG to understand the details, so I am more interested in the statement of the "real"-life problem and the non-trivial result of its mathematical modelling using concepts in algebraic geometry.

Thank you so much! I really appreciate your help. I'm really trying to do some soul searching here and you could really help me with it.

Here's an example of a ``real-life'' application of algebraic geometry.  Consider an optimal control problem that adheres to the Karush-Kuhn-Tucker criteria and is completely polynomial in nature (being completely polynomial is not absolutely necessary to find solutions, but it is to find a global solution).

One can then use the techniques of numerical algebraic geometry (namely homotopy continuation) to solve this system of (nonlinear) polynomial system, find all the complex solutions, throw out any that have ``too large'' of an imaginary part, attain all the real solutions, and check for the optimal one.

A number of software packages exist that can do this (HOMPACK, Phcpack, HOM4PS2.0, POLYSYS_GLP, POLYSYS_PLP).

Some other real-world applications include (but are not limited to) biochemical reaction networks and robotics / kinematics.

These ideas start with Davidenko (50's) and then greatly improved independently by (Drexler) and (Garcia and Zangwill) (late 70's).

Application of algebraic geometry is popular nowadays:

http://meetings.siam.org/program.cfm?CONFCODE=AG17&_ga=2.48674598.1478910260.1545340527-1508289771.1542665618

There are certain successes in coding theory. Numerical tools are giving us an edge to be able to compute things that were not possible before. There is also a growing literature on optimization of polynomial objective functions. All of these can be used for real life problems up to their limitations - nature seems to be always more complex than the math we can compute :) -


What is positive-0 squared minus positive-0? I've got a basic limit problem that I think I'm solving the right way, but I've run into something that looks confusing enough to make me wonder if I'm doing it right.

$$
\lim_{y\to0} \frac{1}{y^2-y} + \frac{1}{y}
$$

I tried approaching this by looking at the Left Hand Limit and Right Hand Limit. Though when I plug in $0^+$, I get

$$
\lim_{y\to0} \frac{1}{0^+-0^+} + \frac{1}{0^+}
$$

What is $0^+-0^+$? Is it still positive? or is it just 0, or in this case, would it be negative? I'm leaning toward negative because for all $0\lt y\lt 1$, $y^2$ will be less than $y$, so then wouldn't I have an infinitely small negative number as a result?

And then, if that turns out to be true, it would appear that this problem is unsolvable by that method then, seeing as I'd have $+\infty + -\infty$, which is undefined.

Clearly $y^2-y\to 0$ as $y\to 0$, as does $y$, so both fractions are blowing up, and their difference is an indeterminate form. Combine them over a common denominator; if you do the algebra correctly, you’ll be able to simplify the resulting fraction to one whose limit as $y\to 0$ is easy to compute.

It's $-1$, silly!  Obviously, you're very confused based on your argument.  I suggest coming back at the problem with fresh eyes after you've taken a break from it.


What is the necessary condition for ODE to have unique solution? For the ODE:
\begin{align}
\dot{x}(t)&=f(x,t) \\
x(t_{0})&=x_{0}
\end{align}
If $\;\;f:\mathbb{R}^{n}\rightarrow{}\mathbb{R}^{n}$ is Lipschitz continuous on $\mathbb{R}^{n}$, then there exists a unique solution for the ODE.
The reverse is not always true, right? and what is the necessary condition for the uniqueness for the ODE?

I have read a book about numerical computing for ODE. The author only assumed there exists a unique solution for the ODE. In proving some properties in the numerical scheme, the author used the Lipschitz continuous property of the ODE. I wonder it is not a correct proof.

The Lipstcitz condition is not necessary for uniqueness of solutions. There are other sufficient conditions, like the Osgood criterion, or monotony conditions like
$$
(f(t,x)-f(t,y)\cdot(x-y)\le0.
$$
As far as I know, there are not known necessary and sufficient conditions.

The first question is answered by Julián Aguirre. The second is not, justifying this addition:

Yosie's theorem of 1926, as presented on p.82 of [1] at least, provides a characterisation of such ODEs' uniqueness, and so a necessary condition as well as sufficient. This uniqueness theorem, however, does not look like many (actually any, to my knowledge) of the classical ones (also presented in [1]). This is because the classical ones (by definition, if you like) boil down to a spatial regularity condition like Lipschitz's or Osgood's. Yosie's theorem, on the other hand, necessarily covers functions which have no meaningful spatial regularity. There are indeed examples of uniqueness for such `spatially irregular' ODEs, as demonstrated quite generally by the elegant idea of [2] to invert the spatial and temporal components in any classical result.

[1] R P Agarwal & V Lakshmikantham, Uniqueness and Nonuniqueness Criteria for Ordinary Differential Equations, DOI: https://doi.org/10.1142/1988

[2] J A Cid & R L Pouso, Does Lipschitz with Respect to x Imply Uniqueness for the Differential Equation y' = f(x,y)?, URL: https://www.jstor.org/stable/27642664


The infinite sum problem How to calculate the following sum

$$1 + (4/7) + (9/7^{2}) + (16/7^{3}) + (25/7^{4}) + \ldots$$

I dont have prior experience to deal with kinds of problem.

Hint: For $|x|<1$ $$\sum_{k=0}^{\infty}x^k=\frac{1}{1-x}$$Differentiate to get: $$ \sum_{k=0}^{\infty}kx^k=\frac{x}{(1-x)^2}$$ Differentiate once again to obtain $\sum_{k=0}^{\infty}k^2x^k$: $$\sum_{k=0}^{\infty}k^2x^k=\frac{x(1-x+x^2)}{(1-x)^3} $$ Here $|x|=1/7<1$, so we can use the above result.


Prove the lecturer is a liar... I was given this puzzle:

  At the end of the seminar, the lecturer waited outside to greet the attendees. The first three seen leaving were all women. The lecturer noted " assuming the attendees are leaving in random order, the probability of that is precisely 1/3." Show the lecturer is lying (or badly mistaken).

I've puzzled it out to proving that there is no ratio of $\binom{a}{3}/\binom{a+b}{3}$ that is 1/3, where $ a,b
\in\mathbb{N}$ and $a\ge3$ and $b\ge0$, $a$ being the number of women and $b$ the number of men.

I'm stuck at this point (but empirically pretty convinced).

Any help/pointers appreciated.

Rasher

PS- as an amusing aside, the first 12 values in the sequence of values for $\binom{3+b}{3}$ are the total number of gifts received for each day of the "12 days of Christmas" song.

I've narrowed it down to proving that in the sequence generated by $n^3+3 n^2+2 n$ with $n
\in\mathbb{N}$ and $n\ge1$ it is impossible for $3(n^3+3 n^2+2 n)$ to exist in the form of $n^3+3 n^2+2 n$ . Still stymied at this point.

I found today a (somewhat) similar question at MathOverflow. Since my question seems to boil down to showing the Diophantine $6 a - 9 a^2 + 3 a^3 - 2 b + 3 b^2 - b^3=0$ has no solutions for $(a,b)
\in\mathbb{N}$ and $(a,b)>= 3$ would it be appropriate to close this here and ask for help at MathOverflow to determine if this can be proved?

An update: I asked a post-doc here at Stanford if he'd have a look (he's done some heavy lifting in the area of bounds on ways $t$ can be represented as a binomial coefficient). To paraphrase his response "That's hard...probably beyond proof in the general case". Since I've tested for explicit solutions to beyond 100M, I'm settling with the lecturer is lying/mistaken at least in spirit unless one admits lecture halls the size of a state.

Let $a$ = the number of women, $b$ = the number of men, and $n = a + b$ be the total number of attendees.

The probability that the first 3 students to leave are all female is $\frac{a}{n} \cdot \frac{a-1}{n-1} \cdot \frac{a-2}{n-2}$.  Setting this expression equal to $\frac{1}{3}$ and cross-multiplying gives $3a(a-1)(a-2)=n(n-1)(n-2)$.

The product of any three consecutive integers is divisible by 6, so the left-hand side is divisible by 18.  For the equation to work out, we must have $n \in \{0, 1, 2\}$ modulo 9.

This doesn't solve your puzzle, but it does rule out (informally) 2/3 of the domain.

Your ratio is $\frac{a! (a+b-3)! 3!}{(a-3)!3!(a+b)!} = \frac{(a+b-3)!}{(a-3)!(a+b)!} = 1/\binom{a+b-3}{a+b},$ so you are trying to prove that $\binom{a+b}{a+b-3}$ is never equal to $3.$ If you look at Pascal's triangle, you will see that the third column is increasing, so you need only check it for a couple of initial values.


Find the value of complex expression $\left(\frac{\sqrt{3}+i}{2}\right)^{69}$ Find the value of

  $$\left(\dfrac{\sqrt{3}+i}{2}\right)^{69}.\DeclareMathOperator{\cis}{cis}$$

I tried to solve this complex expression by converting it into polar form.
I expressed it in polar form $r\cis(t)$ from rectangular form $x+iy$ where $\cis(t) = \cos(t) + i\sin(t)$.
But I am unable to solve further due to the exponent of 69!


What is the difference between equation and formula? Sometimes equation and formula are used interchangeably, but I was wondering if there is a difference.

For example, suppose we can calculate a car's fuel efficiency as:

mpg = distance traveled in miles / the fuel used in a gallon

Is that an equation or formula?

An equation is any expression with an equals sign, so your example is by definition an equation.  Equations appear frequently in mathematics because mathematicians love to use equal signs.

A formula is a set of instructions for creating a desired result.  Non-mathematical examples include such things as chemical formulas (two H and one O make H2O), or the formula for Coca-Cola (which is just a list of ingredients). You can argue that these examples are not equations, in the sense that hydrogen and oxygen are not "equal" to water, yet you can use them to make water.

Mathematicians have long since realized that when it comes to numbers, certain formulas can be expressed most succinctly as equations.  For example, the Pythagorean Theorem $a^2+b^2=c^2$ can be thought of as a formula for finding the length of the side of a right triangle, but it turns out that such a length is always equal to a combination of the other two lengths, so we can express the formula as an equation.  The key idea is that the equation captures not just the ingredients of the formula, but also the relationship between the different ingredients.

In your case, "mpg = distance/gallons" is best understood as "a formula in the form of an equation", which means that in this instance the two words are interchangeable.

One way to answer this question has been developed in the first and second courses of U.S. high school algebra.
In the fist course, the following definitions are formally stated in the glossary.

equation: A statement formed by placing an equals sign between two numerical or variable expressions.

For example, $11-7=4$, $5x-1=9$, and $y+2=2+y$ are all equations because they all satisfy the given definition.
The following definition for a formula is also found formally stated in the glossary:

formula: An equation that states a rule about a relationship.

Here are two useful formulas: $A=lw$, the formula for the area of a rectangle; $P=2l+2w$, the formula for the perimeter of a rectangle.
While at the beginning of the the second course, the following sentence summarizes that the relationship is between two or more variables:

A formula is an equation that states a relationship between two or more variables.


How do you solve $5\sin^2(x)+8\sin x \cos x-3=0$? I have tried using compound angle, sum to product, pythag identities but nothing seems to work. I tried turning it into $\sin 2x$ but then I have $\sin^2(x)$ and $\sin 2 x$ together.

Let $s:=\sin x$. The equation is
$$5s^2\pm8s\sqrt{1-s^2}-3=0$$
or after transformation,
$$64s^2(1-s^2)=9-30s^2+25s^4.$$
The solutions of the biquadratic equation are
$$s=\pm\sqrt{\frac{47\pm8\sqrt{22}}{89}}.$$


Regional Mathematics Olympiad(RMO-India) Geometry Problem How to do this problem? I drew the figure according to the given details but, I believe some extra lines should be drawn to solve this problem.

Applying Menelaus to triangle $ADC$ and transversal $BPF$, we get

$$ \frac{AP}{PD} \times \frac{DB}{BC} \times \frac{CF}{FA} = 1, $$

or that $\frac{ AP}{PD} = 3$.

APplying Menelaus to triangle $AEC$ and tranvsersal $BQF$, we get

$$ \frac{AQ}{QE} \times \frac{EB}{BC} \times \frac{CF}{FA} = 1, $$

or that $\frac{AQ}{QE} = \frac{3}{2} $.

Now, apply Menelaus to triangle $APQ$ and transversal $BDE$ to obtain $ \frac{PB}{BQ}$.  You should be able to take it from here.

Trisect the side of an equilateral triangle will not trisect the corresponding opposite angle. If so, we can trisect any given angle.


Integrate $\int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}$ How to integrate $$\int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}\, dx$$

I believe the exact value of this integral cannot be found , however there could be a way I am not aware of. If not, is it possible to put into a familiar form using other functions such as $Li(x)$ for example.

After some attempts i am still lost and Mathematica is not too helpful.

Thank you kindly for your help and time.

Substitute $x=\frac1{3-2\sqrt2\cos t}$ to get

\begin{align}
& \int_{3-2\sqrt{2}}^{3+2\sqrt{2}} \frac{\ln x}{x\sqrt{-x^2+6x-1}}\, dx\\
= & \int_0^\pi \ln(3-2\sqrt2 \cos t) dt\\
=& \int_0^\pi [\ln2+\ln(1+  \frac12-\frac2{\sqrt2 }\cos t) ]dt=\pi\ln2
\end{align}
where $\int_0^\pi \ln(1+ a^2-2a\cos t)dt=0$ with $a=\frac1{\sqrt2}$ is used.
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$

hint

Put $$x=\frac 1t$$

and observe that
$$\frac{1}{3-2\sqrt{2}}=3+2\sqrt{2}$$

The integral becomes
$$\int_{3-\sqrt{8}}^{3+\sqrt{8}}\frac{\ln(t)}{\sqrt{8-(t-3)^2}}dt$$
Now, we put
$$(t-3)=\sqrt{8}\sin(u)$$
with
$$dt=\sqrt{8}\cos(u)du$$

this yields to

$$\int_{-\frac{\pi}{2}}^{\frac{\pi}{2}}\ln(3+\sqrt{8}\sin(u))du$$


Langton's Ant is unbounded? Suppose I have a grid of black/white squares (not necessarily all white).
How can I prove that Langton's ant is unbounded when it runs on this initial condition?

It seems as though the way to go about this is noting that if it were bounded, it's periodic and achieving a contradiction by looking at the neighbours of the initial square. However, this approach hasn't worked so far for me.

Using dimensionality reduction Turn the lattice into a single dimensional array, iterate the "Langtons ant wave equation" and observe that a) Langtons ant is created and b) $\Psi_\vartriangle$ is indeed a 1D array and look at the variable $k$, a plot of this can be seen in my group on linkedin 104 unbounded highway since this highway is a recurrence relation you can show by telescoping the index $t$ into the index $k$ into the array $\Psi_\vartriangle$ is unbounded in the limit as long as the size of the lattice $E^2$ is taken to the limit of infinity (Calculus).
Graham


Why is one day in a discrete math problem taken as 100,000 seconds? I have the following problem description and I am given several functions:

"What is the largest n for which one can solve within a day using an algorithm that requires f (n) bit operations, where each bit operation is carried out in 10−11 seconds, with these functions f (n)?"

I saw a solution where one day was taken as 100,000 seconds when there are exactly 86,400 seconds. Is such an approach correct? If so, why?

The approximation is not so bad.

Also, what makes you think that a day has 86400 seconds? Some days - where leap seconds occur - have 86401. Consider that the number of 'days in a month' is even less constant.

Save yourself a lot of bother, and pick constants which is are easily defined, and easily divided.


What happens with repeated transformation of $T(x) =x -1/x$? I tested repeated application of $x = x -1/x$ on my computer and could not decide what the long term behavior is.  I iterated up to $500,000$ times and saw variation over a fairly short range from around $-4000$ to $4000$ for starting numbers up to $2,000$.  I realize that limits in the precision of the computer may be a factor. In general, what can be said about the transformation?  Is it an example of chaos?

Just a few observations

Consider the graph $\displaystyle f(x)=x-\frac{1}{x}$. This can be rewritten as $yx=x^2-1$. Since this the discriminant of this conic is positive, we conclude that this graph is a hyperbola. We can find the asymptotes of this hyperbola by considering very large $x$ or very large $y$. In both cases, the $-1$ becomes negligible, so our equation beomes $yx=x^2$, which means either $y=x$ or $x=0$. These two lines are the asymptotes of the hyperbola. For $(0,\infty)$, the graph of the hyperbola is below the line $y=x$. For $(-\infty,0)$, the graph of the hyperbola is above the line $y=x$. So, for any positive number, $x$, we have $\displaystyle x-\frac{1}{x}<x$. Similarly, for any negative number, $x$, we have $\displaystyle x-\frac{1}{x}>x$. The function $f(x)=x-\frac{1}{x}$ is increasing on $\mathbb{R}/\{0\}$. In other words, $0<x_1<x_2 \implies f(x_1)<f(x_2)$ and $0>x_1>x_2 \implies f(x_1)>f(x_2)$.

The graph $f(f(x))$ has $4$ asymptotes: $y=x$, $x=1$, $x=-1$, $x=0$

The graph $f(f(f(x)))$ has $8$ asymptotes: $y=x$, $x=1$, $x=-1$, $x=0$, $x=\frac{1+\sqrt{5}}{2}$, $x=-\frac{1+\sqrt{5}}{2}$, $x=\frac{1-\sqrt{5}}{2}$, $x=\frac{\sqrt{5}-1}{2}$

In general, $f^n(x)$ has $2^n$ assymptotes.

Let's start with a positive value $n$. We know that $f(n)<n$, so if we keep applying the function, $n$ will keep decreasing, until it turns negative. Once it turns negative, if we keep applying the function, $n$ will increase until it is positive, and so on and so forth.

If this transformation converge for some starting point $x$ we would have:

$T(x) = x \rightarrow x - \frac{1}{x} = x \rightarrow \frac{1}{x} = 0$

But then $x = \pm \infty$.

Thus this transformation must diverge or start oscillating.
Easy oscillation that comes to mind is $\pm x$. Then:

$T(x) = -x \rightarrow x - \frac{1}{x} = -x \rightarrow \frac{1}{x} = 2x$

Leaving us with $x = \pm \frac{1}{2} \sqrt{2}$.

Question now remains: for what starting points will we "converge" to this 2 step oscillation and for what starting points will we diverge?
Shortly playing around with it makes it look like only way to oscillate is to start with $\frac{1}{2} \sqrt{2}$ but proof of it hasn't arrived in my head yet.

Another point one might want to address: Are there any other oscillations? So far I only handled the case $T(x) = x$ and partially $T(T(x)) = x$, but what about $T^n(x) = x$? Do those cycle exists?

Edit:
Playing around with it some more I found that numerical errors can kick you out of a cycle quite easily, and you are then put on a magical ride.
Note: $T'(x) = 1 + \frac{1}{x^2}$ is stricly larger than 1 on R so no point is a stable point, meaning that if you make some numerical error (machine precision for example), you will not converge back to the oscilation you were previously on.


Solving equations with factorials? I looked on the internet but couldn't find anything relevant, so I was hoping you could help because I have no clue where to even start with how to solve this equations:
x! = 6

Obviously trial and error here could work, but is there any way to do it for examples where trial and error would take too long?
Thanks.

Using the Stirling approximation

$$n!\approx \sqrt{2\pi n}\left(\frac ne\right)^n$$

or in logarithms,

$$\ln(n!)\approx\frac{\ln(2\pi n)}2+n(\ln(n)-1)=\ln(N),$$

which you can solve for $n$ by numerical methods.

A crude starting approximation is

$$\frac{\ln(N)}{\ln(\ln(N))}.$$

For instance, solving for $N=14!$ yields $n=14.0022249374875\cdots$. No so bad.

The sequence of factorials is strictly increasing, so there is at most one solution.

$3!=6$, so $x=3$ is the only solution.


An intuitive vision of fiber bundles In my mind it is clear the formal definition of a fiber bundle but I can not have a geometric image of it. Roughly speaking, given three topological spaces $X, B, F$ with a continuous surjection $\pi: X\rightarrow B$, we "attach" to every point $b$ of $B$ a closed set $\pi^{-1}(b)$ such that it is homeomorphic to $F$ and so $X$ results a disjoint union of closed sets and each of them is homeomorphic to $F$. We  also ask that this collection of closed subset of $X$ varies with continuity depending on $b\in B$, but I don't understand why this request is formalized using the conditions of local triviality.

Maybe it is helpful to take a map $f\colon\mathbb R\to\mathbb R$ and consider its  graph $\Gamma(f):=\{(x,f(x)) |x\in \mathbb R\}$. We get a continuous map $\pi\colon \Gamma(f)\to \mathbb R$, $(x,f(x)) \mapsto x$. For every $p\in \mathbb R$, the preimage $\pi^{-1}(p)$ is again a single point, so this has a chance to be a fiber bundle. However, it is a fiber bundle, if and only if $f$ is continuous:

If $\pi$ is a fiber bundle near $x$, then there is a local trivialization, i.e. a homeomorphism $\pi^{-1}(U)\to U$ over $U$ for $U\subset \mathbb R$ a neighborhood of $x$. This implies that this homeomorphism is precisely  given by $\pi$ and so the assignment $x\mapsto (x,f(x))\mapsto f(x)$ (which is the inverse of the homeomorphism composed with the projection) is continuous near $x$ and hence $f$ is continuous near $x$. On the other hand, if $f$ is continuous, the map $\pi$ itself is already a homeomorphism (an inverse is given by $x\mapsto (x,f(x))$) and hence a fiber bundle.

So, local triviality in this case means that the map $f$ is locally continuous and the fiber which is essentially $f(x)$, varies continuously with the base point. Also, it is not hard to see that for the sign-function (as an example of a non continuous function), the non continuity at $0$ ruins the local triviality.

I hope this is a little helpful.

One example: A branched cover is a fiber bundle, where the fiber is a set of points.
http://www.math.cornell.edu/~hatcher/AT/ATch1.pdf

See the section "Representing Covering Spaces by Permutations", p. 68.  You can build a branched cover by covering your space with open sets $X \subset \bigcup U_i$  and your bundle with be covered by $U_i \times \{ 1, 2, \dots, n\}$  Then you need to consider what happens on $U_i \cap U_j $.  The transition function will be a bijective map
$$\{ 1, 2, \dots, n \} \to \{ 1, 2, \dots , n \} $$
which is a permutation.

Therefore, branched covers can be thought of as fiber bundles over spaces where the fibers are finite sets.

By covering your space with open sets $X \subset \bigcup U_i$, taking direct products with your fiber $U_i \times V$ and considering what happens over interections $(U_i \cap U_j) \times V$ you can "patchwork" a fiber bundle together.

This construction is very general.  If your fiber $V$ is a vector space, your transitions maps are invertible linear maps
$V \to V$ which take values in the general linear group $GL(V)$.  So the set of vector bundles can be thought of as the space representations of the fundamental group $\pi_1(X) \to GL(V)$.

Then you can ask your transition functions be holomorphic and then it's a holomorphic vector bundle. Or you can ask your transition functions be continuous or have 5 derivatives or "reasonable" restriction (i.e. consistent with vector bundle axioms).

Alternatively, you can study a vector bundle by looking at its section, which are maps from the base space into the vector space.  In high school and college, we deal mostly with the trivial bundle $\mathbb{R}^2$ and where the base space is $\mathbb{R}$ and the fibers are $\mathbb{R}$.  Then we look at sections $$\{ (x, f(x)\} \in \mathbb{R}^2$$
In this way we can consider trivial bundles over the circle $S^1 \times \mathbb{R}$ and consider only those sections which are square-integrable i.e.
$$ \int_{S^1} |f(x)|^2 dx < \infty$$
Morally, this vector bundle is still the cylinder, but we are ruling out certain sections.

Question: What is the analogue of Fourier analysis for the Möbius band in this picture?

The sphere can be thought of as a fiber bundle over the line.  Indeed the fiber
$$ \{ x^2 + y^2 + z^2 = 1 \} \cap \{ z = k\} = \{ (x,y,k): x^2 +y^2 = 1- k^2 \}$$
is a circle except at the endpoints $k = \pm 1$, where the fiber degenerate to points.  Also the torus is a fiber bundle $S^1 \times S^1$.


How to compute $\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi\sqrt 3}$? I came across an old question asking to show that $$\displaystyle\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi}=\frac{\ln 2}{8}.\tag{1}$$
Although I have managed to prove this formula, my proof uses various theta functional relations and looks like an overkill. On the other hand, it suggests a few more identities, for example
$$\displaystyle\sum_{n\text{ odd}}\frac{1}{n\sinh n\pi\sqrt3}=\frac{\ln \left(8-4\sqrt{3}\right)}{4}.\tag{2}$$

Question: can one prove (1) and (2) in a more elementary way? ideally, the proof should make clear further generalizations.

The identities:
$$\sum_{n\geq 1}\frac{(-1)^n}{n^2+m^2} = -\frac{1}{2m^2}+\frac{\pi}{2m\sinh(m\pi)},\tag{A}$$
$$ \frac{1}{m^2+n^2}=\int_{0}^{+\infty}\frac{\sin(nx)}{n}e^{-mx}\,dx,\tag{B}$$
$$ \sum_{n\geq 1}(-1)^n\frac{\sin(nx)}{x}=-\frac{x}{2}+\pi\left\lfloor\frac{x+\pi}{2\pi}\right\rfloor\tag{C} $$
give a wide range of possibilities to evaluate our series. For instance, $(A)$ gives:
$$\begin{eqnarray*}\sum_{k\geq 0}\frac{1}{(2k+1)\sinh(\pi(2k+1))}&=&\frac{1}{\pi}\sum_{k\geq 0}\frac{1}{(2k+1)^2}+\frac{2}{\pi}\sum_{k\geq 0}\sum_{n\geq 1}\frac{(-1)^n}{n^2+(2k+1)^2}\\&=&-\frac{\pi}{8}+\frac{2}{\pi}\sum_{h\geq 1}\frac{r_2(h)\cdot\eta(h)}{h}\tag{1}\end{eqnarray*}$$
where $\eta(h)$ equals $-1$ if $h\equiv 2\pmod{4}$, $1$ if $h\equiv 1\pmod{4}$, zero otherwise, and:
$$ r_2(h) = \#\{(n,k)\in\mathbb{N}^2: h=n^2+(2k+1)^2\}.\tag{2} $$
Now it is well-known that $a^2+b^2$ is the only reduced binary quadratic form of discriminant $-4$, hence:
$$\begin{eqnarray*} \#\{(a,b)\in\mathbb{Z}:a^2+b^2=n\} &=& 4\left(\chi_4 * 1\right)(n)\\&=& 4\left(d_{1(4)}(n)-d_{3(4)}(n)\right) \tag{3}
\end{eqnarray*}$$
so that we can evaluate the RHS of $(1)$ through Dirichlet convolution.

Since we have class number one also in the case $a^2+3b^2$, the situation is almost the same for the other series.

The Mellin transform gives another chance. See, for instance, this related problem.

Don't forget that sinh is an odd function, so we can use
$$
n \, sinh(n \pi) = -n \, sinh(-n \pi)
$$
to reduce the summation to
$$
\sum_{n\in2\Bbb Z+1}\frac{1}{n\,sinh(n\pi)} = \sum_{n\in2\Bbb N+1}\frac{1}{n\,sinh(n\pi)}
$$
which is more directly manipulated (without having to play with summation indices) if written anyway as
$$
\sum_{n=1}^{\infty}\frac{1}{(2n+1)\,sinh((2n+1)\pi)}
$$

Not really sure how to simplify to your end, from here, but perhaps this is helpful anyway.

-Cheers


Right angle triangle simple problem The number of degrees in one acute angle of a right-angled triangle is equal to the number of grades in the other; express both the angles in degrees.

So I have found the following answers :

810/17=47,05... degrees and 810/17=47,05... grades which gives 42,35... degrees

Now, the real answer is the following :

900/19=47,3... degrees and 900/19=47,3... grades which gives 42,63... degrees

The only problem with my answer is the following :

810/17degrees=900/17grades

so : 900/17 grades+ 810/17 grades = 100,5.. grades (but for the rest, everything is fine, I get get 90 degrees perfectly and respect all conditions.)

Would you count this as an error ???

Here's what I did : (Help me see the error)

x degrees= (x+x/9) grades

x grades= (x-x/9) degrees

so..

x degrees+(x-x/9)degrees=90 degrees

17x/9 degrees=90 degrees

17x=810

x=810/17 degrees

By the formulaiton of the problem, we also have 810/17 grades

Conversion

810/17 degrees=(810/17+(810/17)/9) grades=900/17 grades

and

810/17 grades=(810/17-(810/17)/9) degrees= 720/17 degrees

Can somebody tell me where I went wrong ??? (By the way, I see how to obtain the "real" answer, but I don't see why mine would be wrong ...)

Thank you !

The angles of a triangle add up to $180$ degrees, or $200$ grades.

So $0.9 G = D$, where $D$ and $G$ are the measures of an angle in degrees and grades, respectively.

One angle is a right angle, so the other two add up to $90$ degrees, or $100$ grades.  Further, the measure of one in degrees (expressed by $D_i$) is the measure of the other in grades (expressed by $G_i$).

So $G_1 + G_2 = 100$, and $G_1 = D_2 = 0.9 G_2,$ so $1.9 G_2 = 100$ or $G_2 \approx 52.63$ grades.  Then $G_1 \approx 47.37$ grades.

Multiply both by $0.9$ to get $(D_1, D_2) \approx (47.37, 42.63).$

(Where you went off a bit was in your definition of the conversion from degrees to grades.  You have $D = (10/9) G$ and $G = (8/9) D$, neither of which is right.  It should be $0.9 G = D$ or $(0.9)^{-1} D = G$.)

I don't understand all the reasoning of your solution, but you made an error when you changed from $x$ degrees = $x+\frac x9$ grades, which is correct, to $x$ grades = $x - \frac x9$ degrees.  The first says the number of degrees is $\frac 9{10}$ the number of grades.  The second says the number of grades is $\frac 98$ the number of degrees instead of $\frac {10}9$

And yes, if the number of grades adds up to $100.5$ instead of $100$ that is an error.  It should add to exactly $100$.


Calculation of CPU Utilization Problem A computer, has an operating system occupying 1GB of memory. The
  following are the processes with memory sizes and their I/O wait
  percentages: A(512MB, 50%), B(128MB, 60%), C(64MB, 70%). If the goal
  is at least 75% CPU utilization, what would be the minimum system
  memory size requirement? Give examples for how many process A,B, and
  C's should we run at the same time. (Show your formulas and work
  clearly, list all your assumptions if you have any)

The formula for CPU utilization is $1-p^{n}$, in which $n$ is number of process running in memory and p is the average percentage of time processes are waiting for I/O.

$$64 = x, 128 = 2x, 512 = 8x$$

$$0.25 = p^n$$

$$0.25 = 0.5*4x * 0.6*2x * 0.7*x$$

$$0.5 = 16x^3 * 0.42$$

For a limited time, I can't go further. Even if I have time, I don't go further. Is my approach is wrong? Is there any simpler way to find it? It is an exam question.

Given that we have 3 types of processes as follows:
Process A : 512 MB with 50% Idle time
Process B : 128 MB with 60% Idle time
Process C : 64  MB with 70% Idle time

Now we have each of these as follows:
(Number of each of these processes) = x1 of A, x2 of B and x3 of C
We can calculate CPU Utilisation $\eta$ as follows:
$\eta$ = 1 - 0.5^x1 * 0.6^x2 * 0.7^x3
Now we need to play with x1, x2, x3 such as $\eta$ >= 0.75
(x1,x2,x3) = (1,0,2) results in 0.755 which is as close as you can come to 0.75 with x1,x2,x3 integer. -- (1)
with, (x1,x2,x3) = (1,1,1) utilisation comes out to be 0.79, hence result obtained in statement (1) is the closest you can reach atleast 75% CPU utilisation.
Now, with result obtained in (1), let's calculate system memory size M
M = x1* 512MB + x2* 128MB + x3*64MB + 1024MB (OS)
M = 512+2*64+1024 MB = 1664 MB = 1.625GB


Calculate the derivative of a complex norm I'm stuck with a rather trivial looking question. How do you calculate the derivative of the norm of a complex number to it self? Like in

$$
\frac{d|a|^2}{da} = ?
$$

I think it would give rise to a real valued function but in stead I get

$$
\frac{d|a|^2}{da} = \bar{a}
$$

I can't understand this result.

Thanks in advance.

Hint: For $z$ complex

  $$ |z|^2 =z\bar{z}. $$

Note:

  $$ z=x+iy \implies \bar{z}=x-iy .$$

For a complex variable a, |a|^2 derivative exist only at zero. Try evaluating at x_0+i*y_0 not zero. Approach from y=y_0 and you get 2*x_0 but approach from x=x_0 and you get -2*y_0*i. And at zero you get limit r^2 over re^(thetai) that go to zero.


Polygon and line intersection Does anyone help me with the fast algorithm to determine the intersection of a polygon (rotated rectangle) and a line (definite by 2 points)?
The only true/false result is needed.

Take your rectangle $ABCD$, put ,e.g.,
$$
\left| {AB} \right| = b\quad \left| {AD} \right| = d
$$
and consider the unit vectors
$$
{\bf u} = {{\mathop {AB}\limits^ \to  } \over {\left| {AB} \right|}} = {1 \over b}\mathop {AB}\limits^ \to  \quad \quad
 {\bf v} = {{\mathop {AD}\limits^ \to  } \over {\left| {AD} \right|}} = {1 \over d}\mathop {AD}\limits^ \to
$$

Then a generic point $P$ will be inside the rectangle if the projection of $\vec {AP}$
onto $\bf u$ and $\bf v$ is
$$
\left\{ \matrix{
  0 \le \mathop {AP}\limits^ \to   \cdot {\bf u} \le b \hfill \cr
  0 \le \mathop {AP}\limits^ \to   \cdot {\bf v} \le d \hfill \cr}  \right.
$$

The point $P$ will be on and inside the segment if
$$
\mathop {OP}\limits^ \to   = \lambda \mathop {OQ}\limits^ \to   + \left( {1 - \lambda } \right)\mathop {OR}\limits^ \to  \quad \left| {\;0 \le \lambda  \le 1} \right.
$$
or what is the same, if
$$
\mathop {AP}\limits^ \to   = \mathop {AR}\limits^ \to   + \lambda \mathop {RQ}\limits^ \to  \quad \left| {\;0 \le \lambda  \le 1} \right.
$$

So the rectangle and the segment will have points in common iff
$$
\begin{array}{l}
 \left\{ \begin{array}{l}
 0 \le \mathop {AP}\limits^ \to   \cdot {\bf u} \le b \\
 0 \le \mathop {AP}\limits^ \to   \cdot {\bf v} \le d \\
 \mathop {AP}\limits^ \to   = \mathop {AR}\limits^ \to   + \lambda \mathop {RQ}\limits^ \to   \\
 0 \le \lambda  \le 1 \\
 \end{array} \right.\quad  \Rightarrow \quad \left\{ \begin{array}{l}
 0 \le \lambda \mathop {RQ}\limits^ \to   \cdot {\bf u} + \mathop {AR}\limits^ \to   \cdot {\bf u} \le b \\
 0 \le \lambda \mathop {RQ}\limits^ \to   \cdot {\bf v} + \mathop {AR}\limits^ \to   \cdot {\bf v} \le d \\
 0 \le \lambda  \le 1 \\
 \end{array} \right. \\
  \Rightarrow \quad \left\{ \begin{array}{l}
 \frac{{\mathop {RA}\limits^ \to   \cdot {\bf u}}}{{\mathop {RQ}\limits^ \to   \cdot {\bf u}}} \le \lambda  \le \frac{{\mathop {RB}\limits^ \to   \cdot {\bf u}}}{{\mathop {RQ}\limits^ \to   \cdot {\bf u}}} \\
 \frac{{\mathop {RA}\limits^ \to   \cdot {\bf v}}}{{\mathop {RQ}\limits^ \to   \cdot {\bf v}}} \le \lambda  \le \frac{{\mathop {RD}\limits^ \to   \cdot {\bf v}}}{{\mathop {RQ}\limits^ \to   \cdot {\bf v}}} \\
 0 \le \lambda  \le 1 \\
 \end{array} \right. \\
 \end{array}
$$

Therefore it is just to check if the tree intervals specified for $\lambda$
have a common intersection or not, and if they have that will
identify an interval for $\lambda$ wich gives the points of the segment internal
to the rectangle.
Only it is to pay attention to the special cases in which $\mathop {RQ}\limits^ \to   \cdot {\bf u} =0$,
which means that the segment $RQ$ is normal to side $AB$, or otherwise that $\mathop {RQ}\limits^ \to   \cdot {\bf v} =0$
which means normal to the $AD$ edge.
In one of this cases (cannot be both) the corresponding condition shall become , e.g. for the first
$$
\mathop {RA}\limits^ \to   \cdot {\bf u} \le 0 \le \mathop {RB}\limits^ \to   \cdot {\bf u}
$$
and the intersection for $\lambda$ will be given by other two intervals only.

If your lines and rectangles are fairly predictable, a fast techinque is to use the rectangle as aligned to the axis, and rotate the line segment accordingly.

Parameterize the line, and see if any point lies inside the rectangle.


In how many ways can $n$ toys be distributed to $n$ children so that exactly one child does not receive a toy? $n$ toys are to be distributed amongst $n$ children. The total number of ways in which these toys can be distributed so that exactly one child gets no toy is equal to?

I tried it with the concept that if one child gets no toy, then we are left with $n$ toys to be distributed amongst $n-1$ children. But this does not help and the answer is not coming.

Please somebody,can you give the detailed explanation how to solve this sum. Thanks

The question is stated as follows:

  If $n$ $\color{red}{different}$ toys have to be distributed among $n$ children, then the total number of ways in which these toys can be distributed so that exactly one child gets no toy is equal to?

Source: link

Step 1: There are $n$ ways to choose the unlucky child.

Step 2: There are ${n\choose 2}$ ways to choose $2$ toys. Note: only 1 child can get maximum $2$ toys, otherwise more than $1$ child will get no toy.

Step 3: There are $n-1$ ways to choose the luckiest child, who will get the $2$ toys.

Step 4: There are $(n-2)!$ ways to distribute the rest $n-2$ toys to the rest $n-2$ children.

Hence:
$$n\cdot {n\choose 2}\cdot (n-1)\cdot (n-2)!=n!{n\choose 2}.$$

$1$ child gets $0$ toys, $(n-1)$ children get $1$ toy each, any of the $(n-1)$ children can get a $2$nd toy.

$1$ child can be left without a toy in $n$ ways. The 'extra' toy can be chosen in $n$ ways and so can be distributed among the remaining $(n-1)$ children in $n(n-1)$ ways. The remaining $(n-1)$ toys can be distributed among the remaining $(n-1)$ children (one toy each) in $(n-1)$! ways.

However, if the child getting $2$ toys gets toy $A$ as the 'extra' toy and toy $B$ as the 'ordinary' toy, this is the same as if this child gets toy B as the extra toy and toy $A$ as the ordinary toy. So we have counted $2\times$ as many combinations as we need.

So the number of distributions is
$$  \frac{n  n(n-1)(n-1)! }{ 2}
= n!  \frac{n(n-1)}{2}
= n! \frac{ n!}{2!(n-2)!}
= n!  C(n,2). $$


How do I find the coefficient of restitution COR when a sphere is hanging from a roof? The problem is as follows:

The diagram from below shows a sphere labeled $A$ and is moving with a horizontal speed of $v=4\sqrt{10}\,\frac{m}{s}$ over a frictionless table. After the collision the cable makes an angle of $53^{\circ}$ with the vertical and the sphere $A$ ends at rest. Find the $COR$ (coefficient of restitution) and the relationship between the masses.

The alternatives given are as follows:

$\begin{array}{ll}
1.&\frac{1}{2},\, \frac{1}{2}\\
2.&\frac{1}{3},\, \frac{1}{6}\\
3.&\frac{1}{2},\, \frac{1}{3}\\
4.&\frac{1}{8},\, \frac{1}{2}\\
5.&\frac{1}{2},\, \frac{1}{6}\\
\end{array}$

When it mentions the COR for the collision I'm assuming that it is an inelastic collision. Therefore if treated as such then

$p_1=p_2$

$m_1 v_1=m_2 v_2$

$v_2=4\sqrt{10} \frac{m_1}{m_2}$

But that's how far I went. I'm stuck with this problem, can someone help me here?.

Initial momentum of $A$ is $4\sqrt{10}m_A$. This is equal to the momentum of $B$ after the collision. The kinetic energy of $B$ after the collision is then $p^{2}/2m_B = 80m_{A}^{2}\big/m_B$.

As has been noted, $B$ is raised by $2\text{m}$ by the collision, giving it a potential energy of $20m_{B} = \frac{1}{2}m_{B}v_{B}^{2}$. Note therefore that $v_{B} = \sqrt{40} = 2\sqrt{10}$.

Therefore $80m_{A}^{2} = 20m_{B}^{2}$, meaning $m_{A}/m_{B} = 1/2$.

The COR is then $$\frac{v_{B}}{v_A} = \frac{2\sqrt{10}}{4\sqrt{10}} = 1/2.$$

Total mechanical energy using $g=10$ and $\cos(53)=\frac 35$ ...
$$E = \frac 12 m_a (4\sqrt {10})^2=80m_a
\\E'= m_bg[5(1-\cos(53))] = 20 m_b$$
$E'$must be equal to the total Kinetic energy immediately after collision so .
$$C_r =\sqrt{ \frac{KE'}{KE}}= \frac 14 \frac {m_b}{m_a}   $$
The only option that could possibly make sense would be the one in which $C_r=\frac 12$ which must have $m_b = m_a$
It is not entirely clear what they mean by "the relationship between the masses" , it is possible that they mean $\frac{m_b}{m_a+m_b}$, in which case you should choose the option with both numbers equal to $\frac 12$


Sum of digits of sum of digits of sum of digits of $7^{7^{7^7}}$ On the back of a mathematical magazine, I came across some "quick facts" about the number 7. Most of them were real life related ones, like "Rome was buit on 7 hills", "the neck of most mamals is made out of 7 bones", "a ladybird has 7 black spots on its back" etc.

But the last one was very intriguing: denote by $A$ the sum of digits of the number $7^{7^{7^7}}$, by $B$ the sum of digits of $A$, and by $C$ the sum of digits of $B$. What is the sum of digits of $C$?

After doing some research on the power of usual computers, I concluded that it would not be possible to calculculate this number and find the sums like this. So the solution must be pure mathematical. But I could not manage to find any path to a solution. If it helps, we have that $7^7=823543$ which has sum of digits $25$, but online big number calculators can't even compute $7^{7^7}$, and I figured it was not worth it to write such a program myself.

Edit: After some good hours I want to draw some conclusions about the question as well as proposing some extensions.

We saw that for the proposed number $^4 7$, the searched sum is $7$. Mees de Vries found a rigorous proof for that, but which doesn't seem to work in a general case; Henry gave a quite tedious solution which may be generalizable with some extra explanations needed, and Gottfried Helms came up with a computer algorithm which he used to deduce that the sum of digits (repeated as many times as necessary) will repeat with period 3 across consecutive powers of 7.

Now, based on the fact that from Euler's Theorem we will get that $^n 7$ has residue $7$ mod 9, for any $n\in\mathbb{N}^*$, I came to believe that if we repeatedly do sum of digits for larger $^n 7$ we will also get 7. The question is  how many times we would have to do the sum-of-digits. Also, I would like to notice that the number 7 here is not a simple coincidence. It is because it is $\varphi(9)+1$, and the sum-of-digits conserves mod 9. It would be interesting to find a similar property for 3, which has totient function 2, and also "conserves" the residue of the sum-of-digits (here it seems that the sum will be 9 - at least this is how it is for $3^3$ and $3^{3^3}$).

Thanks for all your answers!

Let $\mathrm{sod}$ stand for the upper-bounded sum-of-digits operation, that is,
$$
\mathrm{sod}(n) = \sup_{m \leq n}\left( \text{sum of digits of $m$}\right).
$$
This has the nice advantage that it is trivially a monotone function.

If we can prove that
$$
\mathrm{sod}^4\left(7^{7^{7^7}}\right) < 10,
$$
we can use Euler's theorem to quickly conclude, because applying the (ordinary) sum of digits operation does not change the residue modulo 9. But let's let WolframAlpha do the work for us, to see that the result will be 7.

Note that $\mathrm{sod}(n) \leq 9(\log_{10}(n) + 1)$, as $\log_{10}(n) + 1$ is an upper bound on the number of digits in the number $n$, and each digit is at most 9. Let's focus on the case where $n$ has at least two digits, so that in fact this is upper bounded by $18\log_{10}(n)$. Then we have that
$$
\mathrm{sod}^4\left(7^{7^{7^7}}\right) \leq \mathrm{sod}^3\left(7^{7^7} \cdot 18 \cdot \log_{10}(7)\right) \leq \mathrm{sod}^2\left(7^7\cdot 18 \cdot \log_{10}(7) + 18\log_{10}(18) + 18\log_{10}^2(7)\right).
$$
(Note that we are using the monotonicity of our $\mathrm{sod}$ function here! This is why we made the definition.)

The number in the argument of $\mathrm{sod}^2$ is small enough that we can just evaluate it for simplicity: it is certainly upper bounded by 12527573. The $\mathrm{sod}$ of this number is the sum of digits of $9999999$, which is $7 \times 9 = 63$, and we are left just barely (!) able to conclude, because the $\mathrm{sod}$ of this number is the sum of digits of $59$, which is $14$. Despite the fact that this is larger than the 10 we claimed we needed above, we are still done: if the sum of digits we were after were greater than 7 it would have to be at least 16, and it cannot be, thus we conclude that the result is indeed 7.

$$
\begin{split}
7^{7^{7^7}}= 7^{7^{823543}} &= 7^\text{a number with around $>100000$ but less than $1000000$ digits}\\
&= \text{a number with around a number with 5 digits}
\end{split}
$$
Therefore, $C$ has $5$ digits


How to calculate the sum of that probability distribution? Let $W$ be the random variable that counts the number of tails before one gets $r$ heads for a coin whose probability of heads is $θ$. Without using moment generating function, show that the mean and variance for $W$ are $[r(1-θ)]/θ$ and $[r(1-θ)]/θ^2$. I know it is a negative binomial distribution. But without using mgf, how to calculate that? I tried and I gave up in the half way because it is too complicated.

$W$ is the sum of $r$ independent random variables $W_1,\ldots,W_r$ each of which counts the number of tails before one gets one head.  Therefore the mean and variance are respectively $r$ times the mean and variance of any of those independent random variables.

So $\Pr(W_1=k)=(1-\theta)^k\theta$ for $k=0,1,2,3,\ldots\,{}$.  One way to find $\operatorname{E}(W_1)$ is just applying the definition:
\begin{align}
\operatorname{E}(W_1) & = \sum_{k=0}^\infty k\Pr(W=k) = \sum_{k=0}^\infty k(1-\theta)^k\theta \\
& = \sum_{k=0}^\infty \sum_{j=1}^k (1-\theta)^k \theta \quad (\text{This last sum has $0$ terms when $k=0$.}) \\
& = \sum_{j=1}^\infty \sum_{k=j}^\infty  (1-\theta)^k \theta \\
& = \sum_{j=1}^\infty \frac{(1-\theta)^j\theta}{1-(1-\theta)} = \sum_{j=1}^\infty (1-\theta)^j = \frac {1-\theta} \theta.
\end{align}

To find $\operatorname{E}(W_1^2-W_1)$ is algebraically a bit simpler than to find $\operatorname{E}(W_1^2)$ directcly:
\begin{align}
\operatorname{E}(W_1^2-W_1) & = \sum_{k=0}^\infty k(k-1)(1-\theta)^k\theta \\
& = \theta(1-\theta)^2\sum_{k=2}^\infty k(k-1)(1-\theta)^{k-2} \\
& = \theta(1-\theta)^2\sum_{k=2}^\infty k(k-1)\eta^{k-2} \\
& = \theta(1-\theta)^2\sum_{k=2}^\infty \frac{d^2}{d\eta^2} k\eta^{k-2} \\
& = \theta(1-\theta)^2 \frac{d^2}{d\eta^2} \sum_{k=2}^\infty \eta^k \\
& = \cdots
\end{align}
The last series is geoemtric.  Add it up; do the differentiation; then replace $\eta$ with $1-\theta$.  Then by using the known value of $\operatorname{E}(W_1)$, you can find $\operatorname{E}(W_1^2)$ and then $\operatorname{var}(W_1)=\operatorname{E}(W_1^2)-(\operatorname{E}(W_1))^2$.

Two issues:

That $\sum\dfrac{d}{d\eta}\cdots=\dfrac{d}{d\eta}\sum\cdots$ is easily shown when the sum has finitely many terms.  That it works with infinitely many terms is problematic.  A well-known theorem that's not that hard to prove says it does work when the sum is a power series and one is in the interior of the circle of convergence.
Perhaps it is inelegant for me to use two very different methods for the two expected values.  I have a preference (on this particular occasion of this particular posted question) for the first method because it avoids the issue in the first bullet point above.  (The second method might be somewhat more generally applicable and thus preferable in some other contexts.)  My excuse is that there are only 24 hours in a day and I must attend to some other things.


Domain of function $g$ question when $g(x) = f(x+3)$ Suppose $f$ is defined on $[0;4]$ and  $g(x) = f(x+3)$ what is the domain of $g$?

I'm having a tough time solving this problem. I know that $0 \leq x \leq 4$ for the domain of $f$. So would it be correct to say that for the function $g$, $0 \leq x +3\leq 4$ so:$$f(0) \leq f(x+3) \leq f(4)$$
$$ \Rightarrow f(0) \leq g(x) \leq f(4)$$

I'm not sure how to proceed. A little assistance would be greatly appreciated.

As you say, you require the argument of $f$ to be in $[0,4]$, so

$$x+3\in [0,4]$$
$$\iff 0\leq x+3\leq4$$
$$ \iff (0)-3\leq (x+3) - 3 \leq 4-3$$
$$ \iff -3 \leq x \leq 1$$
$$\iff x\in [-3,1]$$

So the domain of $g$ (the set of valid values of $x$) is $[-3,1]$.

Notice that my approach is "if I know where $x+3$ is, where must $x$ be?".

$$x\in\text{dom(g)}\iff x+3\in\text{dom(f)}\iff x+3\in[0,4]\iff x\in[-3,1].$$


Extension of formula for solution of heat equation on distributions Consider a heat equation
$$
   u_{t} = u_{xx}+f(t,x),\; (x,t) \in (0,L)\times(0,T] \\
   u(0,x) = 0, \; x \in [0,L] \\
   u(t,0) = 0, \; t \in [0,T] \\
   u(t,L) = 0, \; t \in [0,T].
$$
If $f(t,x)$ is a regular function then solution of this system is given by [Tikhonov, Equations of Mathematical Physics]:
$$
   u(t,x) = \int\limits_{0}^{t} \int\limits_{0}^{L} G(x,\xi,t-\tau)f(\xi,\tau)d\xi d\tau,
$$
where
$$
   G(x,\xi,t) = \frac{2}{l}\sum\limits_{n=1}^{\infty} \sin \left( \frac{\pi n}{L}x \right) \sin \left( \frac{\pi n}{L} \xi \right) \exp \left( -\left( \frac{\pi n }{L} \right)^2 t \right)
$$
Tikhonov writes that this formula for solution may be extended to larger class of functions $f$. My question is how large is this class? Which class of distributions we can put in formula for solution? Is it possible for such distributions as $f(t,x)=h(t)\delta(x-x_0)$, where $h$ is a regular function?

The Green's formula given in Tikhonov's book is valid until Green's formula
$$
\int_0^L G f dx
$$
makes sense.

In your case, $G$ is continuous in $x \in [0, L]$, therefore $u(t, x)$ is defined also for $f(x, t) = h(t) \delta(x - x_0)$.

Note that your solution can be obtained by this approach:

Let $u(t,x)=\sum\limits_{n=1}^\infty C(t,n)\sin\dfrac{\pi nx}{L}$ so that it automatically satisfies $u(t,0)=0$ and $u(t,L)=0$ ,

Then $\sum\limits_{n=1}^\infty C_t(t,n)\sin\dfrac{\pi nx}{L}=-\sum\limits_{n=1}^\infty\dfrac{\pi^2n^2}{L^2}C(t,n)\sin\dfrac{\pi nx}{L}+f(t,x)$

$\sum\limits_{n=1}^\infty C_t(t,n)\sin\dfrac{\pi nx}{L}+\sum\limits_{n=1}^\infty\dfrac{\pi^2n^2}{L^2}C(t,n)\sin\dfrac{\pi nx}{L}=f(t,x)$

$\sum\limits_{n=1}^\infty\biggl(C_t(t,n)+\dfrac{\pi^2n^2}{L^2}C(t,n)\biggr)\sin\dfrac{\pi nx}{L}=f(t,x)$

For $0<x<L$ ,

$\sum\limits_{n=1}^\infty\biggl(C_t(t,n)+\dfrac{\pi^2n^2}{L^2}C(t,n)\biggr)\sin\dfrac{\pi nx}{L}=\sum\limits_{n=1}^\infty\dfrac{2}{L}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~\sin\dfrac{\pi nx}{L}$

$\therefore C_t(t,n)+\dfrac{\pi^2n^2}{L^2}C(t,n)=\dfrac{2}{L}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx$

$\dfrac{\partial}{\partial t}\left(e^{\frac{\pi^2n^2t}{L^2}}C(t,n)\right)=\dfrac{2e^{\frac{\pi^2n^2t}{L^2}}}{L}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx$

$e^{\frac{\pi^2n^2t}{L^2}}C(t,n)=\dfrac{2}{L}\int e^{\frac{\pi^2n^2t}{L^2}}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~dt$

$e^{\frac{\pi^2n^2t}{L^2}}C(t,n)=A(n)+\dfrac{2}{L}\int_0^t e^{\frac{\pi^2n^2t}{L^2}}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~dt$

$C(t,n)=A(n)e^{-\frac{\pi^2n^2t}{L^2}}+\dfrac{2e^{-\frac{\pi^2n^2t}{L^2}}}{L}\int_0^t e^{\frac{\pi^2n^2t}{L^2}}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~dt$

$\therefore u(t,x)=\sum\limits_{n=1}^\infty\biggl(A(n)e^{-\frac{\pi^2n^2t}{L^2}}+\dfrac{2e^{-\frac{\pi^2n^2t}{L^2}}}{L}\int_0^t e^{\frac{\pi^2n^2t}{L^2}}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~dt\biggr)\sin\dfrac{\pi nx}{L}$

$u(0,x)=0$ :

$\sum\limits_{n=1}^\infty A(n)\sin\dfrac{\pi nx}{L}=0$

$A(n)=0$

$\therefore u(t,x)=\dfrac{2}{L}\sum\limits_{n=1}^\infty\int_0^t e^{\frac{\pi^2n^2t}{L^2}}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~dt~e^{-\frac{\pi^2n^2t}{L^2}}\sin\dfrac{\pi nx}{L}$

In fact your formula only differ from representation:

$\dfrac{2}{L}\sum\limits_{n=1}^\infty\int_0^t e^{\frac{\pi^2n^2t}{L^2}}\int_0^L f(t,x)\sin\dfrac{\pi nx}{L}dx~dt~e^{-\frac{\pi^2n^2t}{L^2}}\sin\dfrac{\pi nx}{L}$

$=\dfrac{2}{L}\sum\limits_{n=1}^\infty\int_0^t e^{\frac{\pi^2n^2\tau}{L^2}}\int_0^L f(\tau,\xi)\sin\dfrac{\pi n\xi}{L}d\xi~d\tau~e^{-\frac{\pi^2n^2t}{L^2}}\sin\dfrac{\pi nx}{L}$

$=\int_0^t\int_0^L\sum\limits_{n=1}^\infty\dfrac{2}{L}e^{-\frac{\pi^2n^2(t-\tau)}{L^2}}\sin\dfrac{\pi nx}{L}\sin\dfrac{\pi n\xi}{L}f(\tau,\xi)~d\xi~d\tau$

Hence this result works when $f(t,x)$ can be expressed by fourier sine series with respect to $x$ .


Number of Fixed Point(s) of a Differentiable Function For a differentiable function $f:\mathbb R\to\mathbb R$ I need to choose the correct statement(s):

$f'(x)\le r<1~\forall~x\in\mathbb R\implies f$ has at least one fixed point.

$f'(x)\le r<1~\forall~x\in\mathbb R\implies f$ has a unique fixed point.

My attempt:

$g:\mathbb R\to\mathbb R:x\mapsto f(x)-x$ is differentiable on $\mathbb R.$ The search is for the number of roots of $g.$ Now $f'(x)\le r<1~\forall~x\in\mathbb R\implies g'(x)\leq r-1<0~\forall~x\in\mathbb R\implies g$ is decreasing everywhere on $\mathbb R.$ Can I say something from this observation$?$

I can show that if it has fixed point than it has to be unique. Suppose you have two fixed points $x_1<x_2$. Than by MVT there is $\xi$ that $x_2-x_1=f(x_2)-f(x_1) = f'(\xi)(x_2 - x_1 )=r(x_2-x_1)$. So $r=1$ and that is contradiction.

MVT will imply that $|f(x)-f(x')|<|x-x'|, x\neq x'$, all three functions satisfies your conditions but none has fixed points, You need compact metric space to have fixed points other than $\infty$,$f(x)=x+\frac{1}{x},x\in[1,\infty)$, $f(x)=\sqrt{x^2+1},x\in\mathbb{R}$, $f(x)=\log(1+e^x),x\in\mathbb{R}$


Proof by Contrapostive: Any simple examples in First-Order Logic? Does anyone have some simple examples of theorems in FOL that are most easily proven using proof by contrapostive? Every example that I have found so far involves aspects number theory. Any help would be appreciated.

Dan

Taking Carl's tip, say you want to prove that CpCqp.  In one natural deduction system, with proof by contrapositive as a derived rule of inference you can then proceed as follows:

1 |   NCqp hypothesis
2 ||  p hypothesis
3 ||| q hypothesis
4 ||| p 2 repetition
5 ||  Cqp 3-4 conditional introduction
6 ||  KCqpNCqp 1, 5 conjunction introduction
7 |   Np 2-6 negation introduction
8     CNCqpNp 1-7  conditional introduction
9     CpCqp 8 proof by contrapositive

An even shorter example goes:

1 | p hypothesis
2   Cpp 1-1 conditional introduction
3   CNpNp 2 proof by contrapositive


Come up with some fun "equation Limericks" We were discussing "Limericks" in my Calculus class. Specifically, "equation Limericks".

A Limerick is a poem with five lines.

The first, second, and fifth lines should have nine syllables each and rhyme with each other, and the third and fourth should have six syllables each and rhyme with each other.

An obscure subtype of the limerick is the "equation Limerick", which states an equation.

Here are some examples given in class:

A dozen, a gross, plus a score
Plus three times the square root of four
Divided by seven
Plus five times eleven
Is nine squared (and not a bit more).

The integral tee squared dee tee
From one to the cube root of three
Times the cosine
Of three pi over nine
Is the log of the cube root of e.

The lon of e to the four
Times the square root of en twenty-four
Adding six dozen please
Minus eight twenty-three's
Is sixteen, case is closed, shut the door.

I was able to come up with a couple of my own Limericks, but they are a bit simple compared to the ones above.

There is surprisingly not a lot of resources online regarding equation Limericks. Can anyone come up with their own that they would like to share?

WARNING: What you see below is my first-ever attempt at poetry in English.

My take on the classics: $e^{\pi \cdot 2i} = 1$.

We start with the constant called $\pi$ / And then multiply by $2i$ / Apply exponential / (This step is essential) / And one's the result who-knows-why!

Five times the cube root of eight, times three now that is pretty great, plus eight and eleven, the square root is seven, plus eleven is ten and an eight.


Real numbers in math What are real numbers for a person who doesnt know ANYTHING about math, and had to explain them what real numbers are.

Are real numbers only rational and irrational? if so then do we have to say what are rational and irrational

and please it cant have a lot of math involved in it (even if it sounds silly). Like say I am in uni or maybe you as a graduate of a university you want to explain to your niece, your little brother that is what I want like I know the good hard theorems but he wont get it.

So i want to simply break it down for someone who wants to know what a real number actually is

so this is what i have so far

So basically say you are a non math person and this is my explanation would it be valid.
So a real number is basically any number that you can think of, let it be decimal numbers, rational numbers which are just fractions, irrational numbers. Real numbers can range anywhere from $\infty$ to $-\infty$
where $\infty$ is the highest number you can think of in your imagination and $-\infty$ is the lowest number you can think of in your imagination. Like numbers really dont end say the biggest number is $1000000000000000000000000000000000000000$ but there is a number bigger than that like $1000000000000000000000000000000000000000$, or the lowest number you can think of is $-1000000000000000000000000000000000000000$ what about $-2000000000000000000000000000000000000000$, etc real numbers are never ending

If you can write it as a decimal, even if the decimal goes on forever, it's a real number.

If you can write it as a fraction of any two numbers, integers or decimals, or as a power of a positive number (or as a whole power of a negative number), it's a real number.

Real numbers are numbers that are on the real number line,

$x\in\mathbb R \iff -\infty<x<\infty$


Can I find $r$, if $r^m\equiv 1 (\bmod 18)=6$ Recall.

Definition $1$. $Ord_n(a)$ is the smallest number $m$ st. $\quad a^{m} \equiv 1(\bmod n)$
Definition $2$. We say $r$ is a primitive root modulo $n$ if $\operatorname{Ord}_{n} (r)=\phi(n)$
Note: $\left\{1,r, r^2,..., r^{\phi(n)-1}\right\}=U_n$

  I want to find $r$ such that $ord_{18} (r)=\phi(18)=6$, that is we get by definition 1: $r^{m} \equiv 1(\bmod 18)=6$, so hence can I find $r$?

As $(r,18)=1,(r,6)=1$

As $\phi(18)=6,$ord$_{18}r$ must divide $6$

Now $5^2\equiv7\not\equiv1,5^3\equiv-1\not\equiv1\pmod{18}$

$\implies$ord$_{18}5=6$

Finally use

What integers have order $6 \pmod {31}$?


How to find the coordinates of a regular polyhedron We well know that to find the coordinates of the vertices a polygon, we can simply solve the complex variable equation $z^n-1=0$ (and scale them as we wish).

Is there an analog method to find the coordinates of the vertices of a Platonic solid?
For example something like solving an equation such as $(x^2+y^2+z^2)^{n/2}-1 = 0$, for certain numbers $(x,y,z)$? and if so, how to solve this?

Thanks.

H.

When in doubt,  google.

https://en.wikipedia.org/wiki/Platonic_solid#Cartesian_coordinates

It's not a general formula but simply a statement of the 5 cases with graphical explanations why.


Prove alternating group has half of the size of symmetric group Prove: For $n>1$, $A_n$ has order $\frac{n!}{2}$.

The textbook writes the proof as follows:

  For each odd permutation $\alpha$, the permutation $(12)\alpha$ is even and, by the cancellation property in groups, $(12)\alpha \neq (12)\beta$,when $\alpha \neq \beta$. Thus, there are at least as many even permutations as there are odd ones.$\dots$

However, I don't understand why "$(12)\alpha \neq (12)\beta$ when $\alpha \neq \beta$" implies that there are at least as many even permutations as there are odd ones.

$\alpha$ and $\beta$ are odd permutations. Thus the statement says that for every pair of distinct odd permutations $\{ \alpha, \beta \}$, you can find an associated distinct pair of even permutations, namely $\{(12) \alpha, (12) \beta\}$. Therefore, the number of even permutations is greater than or equal to the number of odd permutations.