Using L'Hopital's rule to find the limit of this function I need to find the limit of this function using L'Hopitals rule and I am not sure about my calculations
$$ \lim_{x\to (-\infty)}\frac{e^\frac{1}{x^3}-1}{\arctan\frac{2}{x^4}}$$
Using Lhopital rule and finding the derivative I get
$$ \lim_{x\to (-\infty)}\frac{e^\frac{1}{x^3}*\frac{-3}{x^4} }{\frac{1}{1+\frac{4}{x^8}}* \frac{-8}{x^5}}$$
but Im not sure how to proceed further I tried simplifying but Im not 100% confident that's correct
$$ \lim_{x\to (-\infty)}\frac{e^\frac{1}{x^3}*\frac{-3}{x^4} }{\frac{1}{1+\frac{4}{x^8}}* \frac{-8}{x^5}} = \frac{e^\frac{1}{x^3}*\frac{-3}{1} }{\frac{1}{1+\frac{4}{x^8}}* \frac{-8}{x}} = \frac{e^\frac{1}{x^3}*-3 }{\frac{-8}{(1+\frac{4}{x^8})x}} = \frac{e^\frac{1}{x^3}*-3 }{\frac{-8}{x+\frac{4}{x^7}}} = e^\frac{1}{x^3}*-3  *{\frac{x+\frac{4}{x^7}}{-8}} = 1*-3 {\frac{-\infty+0}{-8}} = -\infty$$

Hint:
We don't actually need L'Hopital's rule
Set $-\dfrac1x=h\implies h\to0^+$
$$\lim_{h\to0^+}\dfrac{e^{-h^3}-1}{\arctan(2h^4)}=\lim_{h\to0^+}\dfrac{e^{-h^3}-1}{-h^3}\cdot\dfrac1{\lim_{h\to0^+}\dfrac{\arctan(2h^4)}{2h^4}}\cdot\lim_{h\to0^+}\dfrac{-h^3}{2h^4}=?$$


Lower semicontinuous function as the limit of an increasing sequence of continuous functions Let $ f:\mathbb{R}^m \rightarrow (-\infty,\infty] $ be lower semicontinuous and bounded from below. Set $f_k(x) = \inf\{f(y)+k d( x,y ): y\in \mathbb{R}^m\} $ , where $d(x,y)$ is a metric. It is easy to see that each $f_k$ is continuous and $f_1  \leq f_2\leq ...\leq f \\$. However, I don't know how to prove that $ \lim_{k \rightarrow \infty}f_k(x) = f(x) $ for every $x\in\mathbb{R}^m $.

Sea $-M$ una cota inferior de $f$ (con $M > 0$).\
Tomemos $x\in X$, para todo $\epsilon > 0$ existe $\delta > 0$ tal que si $y\in B(x,\delta)$ entonces $f(y) > f(x) - \epsilon$.\
          Por otro lado, para todo $k\geq 1$ existe  $y_k \in X$ tal que
          $$f(y_k) + kd(x,y_k) < f_k(x) + \epsilon \leq f(x) + \epsilon$$
         Como $f(y_k) \geq f_k(x) \geq -M$  la desigualdad $kd(x,y_k) < f(x) + \epsilon + M$ es satisfecha. Luego, dividiendo por $k$ se tiene que $0\leq d(x,y_k) < \dfrac{(f(x) + \epsilon + M)}{k}$, como esto sucede para todo $k\geq 1$ deducimos que $d(x,y_k)$ converge a $0$. Luego existe $k_0 \geq 1$ para el cual $d(x,y_k) < \delta$ cuando $k\geq k_0$. De esto, se cumple que $f(y_k) > f(x) - \epsilon$ para todo $k\geq k_0$.\
          Luego $0\leq f(x) - f_k(x) < f(x) - (f(y_k) + kd(x,y_k) - \epsilon) = (f(x) - f(y_k)) - kd(x,y_k) + \epsilon < 2\epsilon$ para todo $k\geq k_0$. Por lo tanto $f_k(x)$ converge a $f(x)$.

This is where the lower semicontinuity comes in. When $y$ is close to $x$, you have a lower bound $f(y)\ge f(x)-\epsilon$. When $y$ is far from $x$, the term $kd(x,y)$ is large. Put these together and you have a lower bound on $f_k(x)$.


Laws of Exponents if base(s) negative In the case where $a,b>0$ and $x,y \in \mathbb R$, we have Laws of Exponents such as these:

$a^xa^y=a^{x+y}$
$(ab)^x=a^xb^x$
$a^{-x}=1/a^x$
$(a^x)^y=a^{xy}$
$a^x/a^y=a^{x-y}$
$(a/b)^x=a^x/b^x$
$(a/b)^{-x}=b^x/a^x$

If instead $a,b<0$, then

Which of the above Laws still hold?
Which don't? (And how can they be modified to become true?)

This seems simple enough to check, so just go through each of the identities so that $a, b < 0$. Let $p = -a$ and $q = -b$, ($p$ and $q$ are clearly positive reals). Then, we can substitute these into the identities:

Start with number 2 because that can be used later.

$(ab)^x = a^xb^x$

Substituting yields $(pq)^x = (-p)^x(-q)^x$. As was stated by J. W. Tanner in the comments, this is slightly meaningless because the left-hand side is a function that has only one output, which is real, and the right-hand side can potentially have multiple and imaginary outputs, specifically when $x$ is fractional with an even denominator or irrational. Therefore, for this one the law doesn't really hold, but could be modified to state that if $x$ is rational with an odd denominator in fully simplified form, then it would hold.

$a^xa^y = a^{x+y}$

Substituting yields $(-p)^x(-p)^y = (-p)^{x + y}$. Since $(-p)^x$ and $(-p)^y$ aren't necessarily uniquely defined, if we make the same assumptions as in the last case for $x$ and $y$ we can see that by rule 2 then we have:
$$(-1)^x(p)^x(-1)^y(p)^y = (-1)^{x+y}(p)^{x+y}$$
Which becomes:
$$(-1)^x(-1)^y(p^{x+y}) = (-1)^{x+y}(p)^{x+y}$$
$$(-1)^x(-1)^y = (-1)^{x+y}$$
At this point, it should be clear that for all of the identities, if we show that the identity holds for $a,b = -1$, it holds for all simply by applying rule 2, (assuming that the $x,y$ satisfy the previously mentioned restrictions. This is true because by applying rule 2 we can break any $(-s)^x$ into $(-1)^x(s)^x$, (again, meeting the restrictions). Therefore, we just need to check all the rules for $-1$.

Although it is not expressed here, quickly checking each of the equations for $a,b = -1$ reveals they are all true, again assuming that $(-1)^x$ is uniquely defined. So now to answer the second question: how can they be changed to work for negative values? Well then the only restriction is that $(-1)^x$ must be uniquely defined. I believe the link in the second comment by J. W. Tanner has more information on that.

Surprisingly, all of them hold. You just have to note when the powers are even because that leads to something like $(-1)^2=1$.

But I guess if you can show these exist of $-1$ and another random negative integer like $-5$ then proof's all good. But I'm leaving $x, y \in \mathbb{N}$ because the rules change magnificently when we walk out of our natural field. I'll show one example later on.

By the way, this is an example that goes to show simple ideas can extend to higher things if only they were defined well. Let's begin.

$(-1)^3 \cdot (-5)^3=-1 \cdot -125 = 125 =(-1 \cdot -5)^3$

$(-1)^{2019}=(-1)^{2018} \cdot -1=(-1)^{2018+1}$

That last one may seem to stem from some contest math but for a number $k \in \mathbb{N}$, $(-1)^{2k}=[(-1)^2]^k$. Boom! The fourth law just popped up. It just shows an even product of (-1)s can be arranged in groups of two of these, which become 1 i.e.

$-1 \cdot -1 \cdot -1 \cdot -1=(-1 \cdot -1) \cdot (-1 \cdot -1)$

This is a very helpful fact to note in future, especially if you ever do number theory. Let's move on.

$(-5)^{-3}=(-1)^{-3} \cdot 5^{-3} = \dfrac{1}{5^3} \cdot \dfrac{(-1)^0}{(-1)^3}=\dfrac{1}{(-5)^3}$

$(-5)^2 \div (-5)=\dfrac{(-1)^2 \cdot 5^2}{-1 \cdot 5}=\dfrac{-1 \cdot 5^{2-1}}{1}=(-5)^1=(-5)^{2-1}$

$\dfrac{(-5)^x}{(-1)^x}=(-5)^x \cdot [(-1)^{-1}]^x=[-5 \cdot (-1)^{-1}]^x=(\dfrac{-5}{-1})^x$

$(\dfrac{-5}{-10})^{-x}=[\dfrac{-5}{-10}^x]^{-1} \qquad \text{Which is the reciprocal of the case that preceded it.}$

So there you have it. Now for that special case.

In some textbooks Law. 2 is converted into something like:

$\sqrt{a} \cdot \sqrt{b}=\sqrt{ab} \quad \text{Where in this case } x=0.5$

Both a and b can't be negative because the equation can be shown to produce two separate numbers. Let's say $-16,-36$

$\sqrt{a} \cdot \sqrt{b}=\sqrt{-16} \cdot \sqrt{-36}=\sqrt{-1} \cdot 4 \cdot \sqrt{-1} \cdot 6=-1 \cdot 24$

$\sqrt{ab}=\sqrt{-16 \cdot -36}=\sqrt{16 \cdot 36}=4 \cdot 6$

One is positive and the other is negative. So that's the only exception. Peace.


Prove $\int_0^{\infty }\frac1{\sqrt{x}}\left(\frac{\cos(\pi x^2)}{\sinh (\pi x)}-\frac1{\pi x}\right)dx=\frac{1}{\sqrt{2}}\zeta(\frac{1}{2})$ I encountered an astonishing integral (numerically verified)

$$\int_0^{\infty } \frac{1}{\sqrt{x}}\left[\frac{\cos \left(\pi  x^2\right)}{\sinh (\pi  x)}-\frac{1}{\pi x}\right] \, dx=\frac{1}{\sqrt{2}}\zeta\left(\frac{1}{2}\right)$$

What technique should we use to establish it? Any help will be appreciated.

Let $$\mathscr{I}(s) = \int_0^\infty  {{x^{s - 1}}\left( {\frac{{\cos (\pi {x^2})}}{{\sinh \pi x}} - \frac{1}{{\pi x}}} \right)dx} $$
It suffices to prove, for $0<\Re(s)<1$, $$\tag{*}\mathscr{I}(s) + 2\frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2}\mathscr{I}(1 - s) = 2\frac{{\Gamma (s)\zeta (s)}}{{{{(2\pi )}^s}}}$$

We need a nontrivial Fourier transform:

(Lemma 1) For $\xi\in \mathbb{R}\setminus \{0\}$, $$\int_{ - \infty }^\infty  {(\frac{{{e^{i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}){e^{ - 2\pi ix\xi }}dx}  = i\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2i{\chi _{(0,\infty )}}(\xi )$$
where $\chi_A$ is the characteristic function of set $A$.

Proof: Let $C(r)$ be a contour along real axis, but with small indention above $r \in \mathbb{R}$, then $$\int_{C(0)} {\frac{{{e^{i\pi {z^2}}}{e^{ - 2\pi iz\xi }}}}{{\sinh \pi z}}dz}  = {e^{ - i\pi {\xi ^2}}}\int_{C( - \xi )} {\frac{{{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}dz} $$
Let $F(z)=\dfrac{{{e^{i\pi {z^2}}}{e^{4\pi z}}}}{{\sinh \pi (z + \xi )\sinh 4\pi z}}$,
then $$F(z) - F(z + 4i) = \frac{{2{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}$$
therefore
$$\int_{C( - \xi )} {\frac{{2{e^{i\pi {z^2}}}}}{{\sinh \pi (z + \xi )}}dz}  = \int_C {F(z)dz}  = 2\pi i\color{red}{\frac{{1 - {e^{\pi \xi }}{e^{i\pi {\xi ^2}}}}}{{\pi \sinh \pi \xi }}}$$
where $C$ is the rectangular contour with vertices $\pm \infty, \pm \infty + 4i$, and has small indentions above $-\xi, -\xi+4i, 0, 4i$. Observe that $F$ has $20$ poles inside $C$, summing over residues at these points gives the red expression. Thus,
$$\begin{aligned}&\int_{ - \infty }^\infty  {(\frac{{{e^{i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}){e^{ - 2\pi ix\xi }}dx}  = \int_{C(0)} {\frac{{{e^{i\pi {z^2}}}{e^{ - 2\pi iz\xi }}}}{{\sinh \pi z}}dz}  - \int_{C(0)} {\frac{{{e^{ - 2\pi iz\xi }}}}{{\pi z}}dz}  \\ &= i\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2i{\chi _{(0,\infty )}}(\xi )\end{aligned}$$
proving the lemma.

Let $0<\Re(s)<1$, we have $$\int_0^\infty  {{x^{s - 1}}{e^{ - 2\pi ix\xi }}dx}  = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}{\xi ^{ - s}}\qquad \Im(\xi)\leq 0$$
Plancherel theorem
$\int_\mathbb{R} f(x) \overline{g(x)} dx = \int_\mathbb{R} \hat{f}(\xi) \overline{\hat{g}(\xi)} d\xi$ produces
$$\begin{aligned}&\int_0^\infty  {{x^{s - 1}}\left( {\frac{{{e^{-i\pi {x^2}}}}}{{\sinh \pi x}} - \frac{1}{{\pi x}}} \right)dx}  = \frac{{\Gamma (s)}}{{{{(2\pi i)}^s}}}\int_{ - \infty }^\infty  {{\xi ^{ - s}}\left[ { - i\frac{{{e^{i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} - 2i{\chi _{(0,\infty )}}(\xi )} \right]d\xi }
\\ &=  - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}{e^{ - \pi is/2}}i\int_0^\infty  {{\xi ^{ - s}}\left[ {\frac{{{e^{i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2} \right]d\xi }  + \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}{e^{\pi is/2}}i\int_0^\infty  {{\xi ^{ - s}}\frac{{{e^{i\pi {\xi ^2}}} - {e^{ - \pi \xi }}}}{{\sinh \pi \xi }}d\xi }\end{aligned}$$
Taking complex conjugation (i.e. replace $i$ by $-i$), then sum up with the original gives:
$$\begin{aligned}\mathscr{I}(s) &=  - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2}\int_0^\infty {{\xi ^{ - s}}\left[ {\frac{{{e^{ - i\pi {\xi ^2}}} - {e^{\pi \xi }}}}{{\sinh \pi \xi }} + 2 + \frac{{{e^{i\pi {\xi ^2}}} - {e^{ - \pi \xi }}}}{{\sinh \pi \xi }}} \right]d\xi } \\
& = - \frac{{\Gamma (s)}}{{{{(2\pi )}^s}}}\sin \frac{{\pi s}}{2} \left[ 2\mathscr{I}(1 - s) + 2\int_0^\infty  {{x^{ - s}}\left( {\frac{1}{{\pi x}} - \coth \pi x + 1} \right)dx}\right]
\end{aligned}$$
the next lemma completes the proof of $(*)$.

(Lemma 2) For $0<\Re(s)<1$, $$\int_0^\infty  {{x^{s - 1}}\left( {\frac{1}{{\pi x}} + 1 - \coth \pi x} \right)dx}  =  - \sec \frac{{\pi s}}{2}\zeta (1 - s)$$

Proof: A lucid method is via Mellin inversion. First, expand $\coth \pi x$ in power series of $e^{-\pi x}$, termwise integration yields
$$\int_0^\infty  {{x^{s - 1}}\left( {1 - \coth \pi x} \right)dx}  =  - {2^{1 - s}}{\pi ^{ - s}}\Gamma (s)\zeta (s) =  - \sec \frac{{\pi s}}{2}\zeta (1 - s) \qquad \Re(s)>1$$
where we used the functional equation of $\zeta$. Note that $\zeta$ has "moderate growth" in every vertical strip, so Mellin inversion is permissible
$$\frac{1}{{2\pi i}}\int_{\sigma  - i\infty }^{\sigma  + i\infty } {\left[ { - \sec \frac{{\pi s}}{2}\zeta (1 - s)} \right]{x^{ - s}}ds}  = 1 - \coth \pi x \qquad \sigma>1$$
Now shift the path of integration, to make it has real part $0<\sigma'<1$, taking residue at $s=1$ into account, the LHS of above equation equals
$$  \frac{1}{{2\pi i}}\int_{\sigma ' - i\infty }^{\sigma ' + i\infty } {\left[ { - \sec \frac{{\pi s}}{2}\zeta (1 - s)} \right]{x^{ - s}}ds}  - \frac{1}{{\pi x}} \qquad 0<\sigma'<1$$
apply Mellin inversion again proves the lemma.


Finding the Order of G, given subgroups I need help with a practice exam question:

$G$ is a cyclic group that has only 3 subgroups: $e$ (identity), $G$, and $G'$ s.t. |$G'$| = $n$. What is |G| if:

a) $n = 5$

b) $n$ is any prime number $p$.

What I know is that the order of $G'$ divides the order of $G$; so, does that mean |$G$| = $5k$ for some integer $k$? If so, is it the same for $p$, where |$G$|  = $pk$? Thanks for the help!

Hint: The cyclic group of order $ n $ has a unique (cyclic) subgroup of order $ d $ for each $ d $ dividing $ n $. If a cyclic group has $ 3 $ subgroups, this would mean that its order has $ 3 $ positive integer divisors. Which positive integers have exactly $ 3 $ positive integer divisors?

I am not sure, but here goes. $G'$(of order 5) is the only proper subgroup of $G$ and assume $|G|= n$. So consider an element $a \in G/G'$. So $a^5 \neq e$. Now, $(a^5)^{n/5} = e$. Denote this element by $b$.Thus $b$ generates another proper subgroup of $G$ distinct from $G'$. Which is not possible. So only case is $n/5=1 \implies n = 5$. Same for any prime $p$. (sorry for the bad notation)


Compute $\int_{\gamma}\frac{\log(1+z)}{ (z-\frac{1}{2})^3}dz$ Compute
$$\int_{\gamma}\frac{\log(1+z)}{(z-\frac{1}{2})^3}dz$$
Where $\gamma:[0,2\pi]\to \Bbb{C}$ is given by $\gamma(t)=\frac{2}{3}e^{it}$.

Im quite confused in which branch of the logarithm should I use. And how should I proceed once I've chosen the correct one?

log(1+z) is analytic if we remove the ray (-inf,-1] and the given curve is in the resulting region. Thus from Cauchy formula f''(a)=2/2.pi.i.integral round C of f(z)/(z-a)^3 we get that the integral is (log(1+z))''.pi.i at z=1/2. It is i.pi.1/(9/4)


how to compute the homology groups of a torus with n points removed? i have really big problems in computing the homology groups and even don't really know how to write the exact sequence for different spaces.
i tried to take the torus with n points removed,homotopic to a circle with n circles attached around and use my intuition to get the answer which i'm not sure if is true...
can someone please give a detailed answer? i really want to learn this subject.

After removing $1$ point, and seeing the torus as a $CW$ complex (given by making the usual identifications $I^2/\sim$ of a square) then before gluing, we can enlarge the hole to the boundary of the square. After making identifications, this gives the wedge of two circles.

One can continue in this way to see that the torus with $n$ points removed gives the wedge of $n+1$ circles. In doing this, we can compute the (reduced) homology with Mayer-Vietoris (which is done in Hatcher, if I remember correctly.)

The easiest way to do this is probably induction on the number of circles, and noting that for the wedge of two circles, you can find neighborhoods so that  $A \cup B=X$ and $A \cap B$ is contractible.

Since you know how to compute its fundamental group, the first homology group is the abelianization of the fundamental group. The zero homology groups is isomorphic to $\mathbb{Z} $ since it is connected. And top homology group is zero since it is not compact.

Top homology of an oriented, compact, connected smooth manifold with boundary


How to measure the angle of the tangent to an ellipse? If we consider the ellipse in the picture here

How do we determine the angle $\lambda$ of the vector v (tangent at point x = 2 ,y = 3) with the line joining the center (10,0)and the point (2,3)?

Edit:

a = 10, b = 5

it seems impossible it is 80°, it looks more like 56-57°

$$ \dfrac{x^2}{a^2}+ \dfrac{y^2}{b^2} =1 $$

Differentiate to find slope

$$ y^{'}= \dfrac{x\,  b^2 }{y\, a^2 }= \dfrac{8\, 5^2 }{3 \,10^2} = \dfrac23 $$

Find arctan of above and add to $ {20.55}^0  $ to get correct angle.

The angle $ \lambda $ is: 90°- 9.46° = 80.54°.

If you look closely you can see that the angle $ \lambda $ is not measured from the line joining the center of the ellipse to the point (the green line), but from the line joining the apopsis to the point (the red line).

The angle of the velocity vector is:
$$  \arctan\left( \frac{ e + \cos(\nu)}{ \sin(\nu)} \right) = 39.44\!° $$

where e is the eccentricity, and $\nu$ is the true anomaly $( e = \frac{ \sqrt{3} }{2} ; \ \ \nu = 92.52\!° )$.

The angle of the the line joining the center [10,0] and the point [2,3] ([10,0]-[2,3]=[8,-3]) is:
$$ \arctan \left( -\frac38 \right) = -20.56 \!° $$
So the angle between that line and the tangent vector is: 39.44° + 20.56° = 60°.


Word problem of Arithmetic Series This question is taken from  my text book, given in the Arithmetic Series exercise.
I don't how to find the days using Arithmetic series?

Q: A besieged fortress is held by 5700 men who have provisions for 60 days. If the garrison loses 20 men each day, for how many days will the  provision last?

Thanks

@zonnie

Each guy eats $\frac{1}{342000}$ of the food per day. So we have the sum

$$\frac{\sum _{k=0}^d (5700-20 k)}{342000}=1$$

The sum in the numerator is easy and the whole thing becomes

$$-\frac{(d-570) (d+1)}{34200}= 1 $$

solving the quadratic we get

$$d = 66.99080479207322, d = 502.0091952079268 $$

The second answer is meaningless so the food should hold out for 66 full days and on the 67th day they are hungry. But since we called the first day 0 ( in the sum ) we should add 1 so they will last 67 days and on the 68th they are out.

I think the assumption being made is that each man eats a constant amount of food each day, call it $r$. The statement that they have provisions for $60$ days is supposed to be interpreted as saying that, if the number of men remained constant, they would last $60$ days at that rate. So they have $60\cdot5700\cdot r$ units of food.

If $x_n$ is the amount of men still alive on day $n$, then on that day they'll eat $rx_n$ units of food. Can you work out the formula for $x_n$ and use that to figure out when they'll have eaten the $60\cdot5700\cdot r$ units they started with?


Geometry problem with two chords originating from a common point outside the circle. This one's got me pulling my hair.

By the power-of-a-point theorem,

$$OA\cdot OB=OP\cdot OQ.$$

Construction: Join BQ and AP

Proof

$1. \angle BQO=\angle PAO$(Angle subtended betwewn a chord and a tangent is equal to the angle subtended by the chord on its alternate segment),

$2.\angle QBO=\angle OPA$(Angle subtended betwewn a chord and a tangent is equal to the angle subtended by the chord on its alternate segment)

SO THEY ARE SIMILAR

$\frac{AO}{OQ}=\frac{OP}{OB}$

$\frac {AO}{12}=\frac {4}{18}$

Solve it to get value of AO.


How do I know an element generates a coordinate ring K[W] as a vector space over K? I have an example which proves that a cuspidal cubic $W\subset \mathbb{A}^2$ defined by $y^2-x^3=0$ is not isomorphic to to $\mathbb{A}^1$. I'll start by defining a few things:

Let $V=\mathbb{A}^1$, $W\subseteq \mathbb{A}^2$ and let $\phi : W \rightarrow W$ be the morphism $t \rightarrow (t^2, t^3)$. We show $V$ and $W$ are not isomorphic by showing $K[W]$ and $K[V]$ are not isomorphic where these are the coordinate rings of $W$, $V$ respectively. It is given that $I[W] =  \langle x^3-y^2 \rangle$ so we have that $K[W]=K[x,y]/\langle x^3-y^2 \rangle$. I understand the overall reasoning of the question except one part:

What I dont understand is the following:
'$\overline{(x^a*y^b)}$ generates $K[W]$ as a vector space over $K$'

What does this mean? I understand $\overline{(x^a*y^b)}$ is a coset and element of $K[W]$ but how does it generate it? And what does it mean by as a vector space over $K$? Thanks!

One can see that these two rings are not isomorphic because $I[W]=K[x]\oplus K[x]y\not\simeq K[z]$ since every time $y^2$ occurs, it can be replaced by $x^3$.    Now, to answer your questions.

Recall that a vector space over a field $K$ is a commutative group $V$ with a addition operation $+:V\times V\rightarrow V$ and a scalar multiplication operation $\ast:K\times V\rightarrow V$.  These operations are associative and distributive.

In the case of $K[x,y]$, the monomials $x^ay^b$ generate the ring as a vector space over $K$ because every polynomial can be written as a sum of scalars times monomials.  In the quotient, these elements still generate because you have the natural surjection
$$
K[x,y]\rightarrow K[x,y]/\langle x^3-y^2\rangle.
$$
The resulting space is also a vector space because you can multiply by scalars (elements of $K$) and get another element of the quotient (such a multiplication actually commutes with the quotient map).

Now, you must check the various vector space properties.  Most of the properties come directly from the fact that $K[x,y]$ is a vector space.


Elements that are their own inverses in a symmetric group. How many elements are their own inverses in $S_6$? I'm having a hard time figuring out how to calculate such a thing.

Here are the key facts:

A permutation is its own inverse iff it has order $2$. (Such permutations are called involutions.)
The order of a permutation is the lcm of the orders of the cycles in its cycle decomposition.

This should give you the type of all possible cycle decompositions.

In S6, here are three types of permutations of order 2: a the transpostions (ij) that swap the digits i and j (for i and j different digits between 1 and 6), the double transpostions (ij)(kl), where i, j, k, l are four different digits between 1 and 6, and the triple transposition (ij)(kl)(mn), where i,j,k,l,m,n are six different digits between 1 and 6.

Obviously the number of transpositions (ij) is 6 . 5 / 2, which is 15.

For the double transpositions, the transformation of order 2 will be $(1/2((6.5)/2)((4.3)/2))$ =45.

For the triple transposition, it will be
$(1/3((6.5)/2)((4.3)/2)((2.1)/2))=30$

And also since identity is self inverse element too
Therefore total number of elements of order 2 in S6 = 15+45+30+1 =91.


Does a long exact sequence of flat modules remain exact after tensoring with an arbitrary module? In Liu's Algebraic Geometry and Arithmetic Curves, Proposition 1.2.6 states that given any short exact sequence $0 \rightarrow M' \rightarrow M \rightarrow M'' \rightarrow 0$ with $M''$ flat, taking the tensor product of this with any module $N$ gives an exact sequence $0 \rightarrow M' \otimes N \rightarrow M \otimes N \rightarrow M'' \otimes N \rightarrow 0$.

Later in the book, in Exercise 5.2.8, page 193, he suggests using this Proposition to show that given a long exact sequence which is eventually $0$:

$$ 0 \rightarrow M_0 \rightarrow M_1 \rightarrow \cdots $$

of flat modules, taking the tensor product with an arbitrary module still gives an exact sequence

$$ 0 \rightarrow M_0 \otimes N \rightarrow M_1 \otimes N \rightarrow \cdots$$

How does this generalization follow? It appears that in the proof of the Proposition, he doesn't really use the surjectivity of $M \rightarrow M''$, but then again, all he proves is that $M' \otimes N \rightarrow M'' \otimes N$ is injective; the full exactness being something that always happens for tensor products of short exact sequences.

In fact, Liu says (on page 193, Exercise 2.8(b)) that the sequence is "zero from a finite rank on". Then you start the breaking from the rightmost side of the sequence: $0\to X_{n-1}\to M_{n-1}\to M_n\to 0$, and so on. You know that $M_{n-1}$ and $M_n$ are flat, so $X_{n-1}$ is also flat, and so on.

let $$0 \rightarrow M_0 \rightarrow M_1 \rightarrow M_2 \rightarrow M_3 \rightarrow \cdots$$ be exact.
then there is exact sequences (break):
$$0 \rightarrow M_0 \rightarrow M_1 \rightarrow C_1 \to 0$$ and $$0\to C_1 \rightarrow M_2 \rightarrow C_2 \to 0$$  and so on...

now tensor with $N$ and join sequences again.


Arrangement of $100$ articles numbered $n_{1},n_{2},n_{3},n_{4} \cdots n_{100}$. There are $100$ articles numbered $n_{1},n_{2},n_{3},n_{4} \cdots n_{100}$
  They are arranged in all possible ways. How many arrangments would be there
  in which $n_{28}$ will always be before $n_{29}$

$a.)\ 5050\times 99! \\
b.)\ 5050\times 98! \\
\color{green}{c.)\ 4950\times 98!} \\
d.)\ 4950\times 99! \\ $

I integrated $n_{28}$ and $n_{29}$ in one term and concluded
that answer will be $99!$.

But that's not in options.

I look for a short and simple way.

I have studied maths upto $12$th grade.

Intuitive and Simple

Use the symmetry of the problem. Half of the cases of all arrangements, which is half of $100!$ should be the answer. This can be written as $\frac{1}{2}\times 99\times100\times 98!$ which is $4950 \times 98!$

Of the total arrangements, half of them have n28 before n29 and other half arrangements have n28 after n29.
So,
  total arrangements =100!
  half of them = 100!/2=50*99!=50*99*98!=4950*98!
hope you understood!


Strategy for board game 2 In this question the following was asked:

  Alice and Bob are playing the following game: They have a $4 \times 4$ empty grid and take turns coloring one square each, starting with Alice, both using the same color. Whoever completes any $2 \times 2$ area on the grid (after having made their move) is the loser. Is there any winning strategy for any of the two players?

The answer was that Bob had a winning strategy (see link). It was also determined that for an $n \times n$ grid, where $n$ is odd, Alice has a winning strategy. However, it was not determined who has a winning strategy when $n$ is even with $n \gt 4$.

Can someone spot such a strategy?

Edit

To avoid repeats of answers previously given, here are two strategies for Bob which don't work:

Bob's winning strategy for $n=4$

If Alice colors $(i,j)$, Bob colors $(1+(i+m-1) \mod n, \ j)$, where $n=2m$. Won't work for $n \gt 4$ as Alice can color e.g. $(1,1)$, $(1,2)$, $(n,1)$, $(n,2)$.
Bob mirror's Alice's move

If Alice colors $(i,j)$, Bob colors $(n+1-i, n+1-j)$. Won't work as Alice can color two adjacent central squares.

In fact, I think Alice might have a winning strategy. I simulated $10,000$ games on a $6 \times 6$ grid where each player made random "legal" moves, i.e. moves which don't immediately result in a loss, and Alice consistently wins $56 \text {%}$ of the time.

Simulating random games tells nothing about whether or not a winning strategy exists.  However, as the maximum number of moves before the game ends results in Bob losing (a maximum of 27 out of the 36 squares can be colored without coloring in any 2x2 square).  As a result, as long as Alice can force the game to go the maximum number of moves, she wins.


$x_0 \in [0,\infty)$ and $x_{n+1} =\sqrt{\frac{3x_n+2}{2}}$. Compute $\lim_\limits{n\to\infty} x_n$ Suppose that $x_0 \in [0,\infty)$ and $x_{n+1} =\sqrt{\frac{3x_n+2}{2}}$. Demonstrate that $ (x_n)_{n \in \mathbb {N}}$ converges and compute  $\lim_\limits{n\to\infty} x_n$.

We've not been taught yet how to use L'Hopital or Taylor expansion or asymplotes; I'm in third year of highschool and we've been taught about limits for half a year now, don't expect a hard way to solve this. It's a problem from my test.

Hint: The sequence converges to $2$.
To guess the limit take $n\rightarrow \infty$ at both sides of the recursion, and assume $x_n$ is converging. Since, $x_n$ converges the LHS is equal to $a$, and the RHS is equal to $\sqrt{\dfrac{3a+2}{2}}$. Solve this equation to find the candidate limits.

Now, for the convergence.

If $x_0\in [0,2)$, show by induction that $x_n<2$ for all $n$. Once you have establish that, show, by induction, that $x_n$ is increasing, hence $x_n$ must be a convergent sequence.

The case $x_0\in(2,\infty)$ can be handed similarly.

Last case $x_0=2$ is the easiest.

Hints:

$\displaystyle x_{n+1} - x_n = \frac{\sqrt{3x_n+2} - \sqrt{3x_{n-1}+2}}{\sqrt{2}} = \frac{3(x_n-x_{n-1})}{\sqrt{2}\left(\sqrt{3x_n+2} + \sqrt{3x_{n-1}+2}\right)}\,$, therefore the differences between consecutive terms have the same sign, so the sequence is monotonic;
$\displaystyle x_{n+1} =\sqrt{\frac{3x_n+2}{2}} \iff 2x_{n+1}^2=3x_n+2=0 \iff 2(x_{n+1}^2-4) = 3(x_n-2)\,$, therefore $x_{n+1}-2$ and $x_n-2$ have the same sign, so all terms are on the same side of $2$;
$\displaystyle x_1 >= x_0 \iff 2x_0^2-3x_0-2 = (x_0-2)(2x_0+1) \le 0 \iff x_0 \le 2\,$, therefore:

for $x_0 \le 2$ the sequence is increasing and bounded above by $2$;
for $x_0 \gt 2$ the sequence is decreasing and bounded below by $2$.

[ EDIT ]   For the sake of the belated downvoter, here is the last hint:

the sequence converges since it is bounded and monotonic, then let $\,x\,$ be its limit; passing the recurrence relation to the limit gives $\,x = \sqrt{\frac{3x+2}{2}} \implies 2x^2-3x-2=0\,$ with $\,x \ge 0\,$.


Countably generated $\sigma$-algebra implies separability of $L^p$ spaces Let $\Sigma = \sigma(\mathcal C)$ be the $\sigma$-algebra generated by the countable collection of sets $\mathcal C \subset \mathcal{P}(X)$. How can I prove that if $\mu$ is a $\sigma$-finite measure on $(X,\Sigma)$ then $L^p(X)$ is separable for $1 \le p < \infty$?

I know that simple functions are dense in $L^p(X)$, so I would like to find a countable subset of the set of simple functions that is dense in them. Could you help me please?

Before going into a formal proof, here is the idea. The space is $\sigma$-finite, so we can "break" it into countably many spaces of finite measure. Up to some technical considerations, we are reduced to the case $X$ of finite measure. An algebra generated by a countable class is countable and we can approximate elements of finite measure by those of a generating algebra.

Let $(A_n,n\geqslant 1)$ be a partition of $X$ into measurable sets of finite measure.

We show that $(A_n,A_n\cap \Sigma,\mu_{\mid A_n\cap \Sigma})$ is separable. Consider $f\in L^p(A_n)$ and fix $\varepsilon>0$. There is $f'=\sum_{j=1}^J a_j\chi_{B_j}$ simple simple such that $\int_{A_n}|f-f'|^p\mathrm d\mu\lt\varepsilon^p$. Define $\mathcal A_n$ the algebra generated by sets of the form $A_n\cap C,C\in\mathcal C$. Then $\mathcal A_n$ is countable. Approximate $B_j$ by $B'_j$, an element of $\mathcal A_n$, that is, such that $\mu(B'_j\Delta B_j)\lt \frac 1{J(|a_j|^p+1)}\varepsilon$. Defining $f'':=\sum_{j=1}^Ja_j\chi_{B'_j}$, we get $\lVert f-f''\rVert^p\lt 2\varepsilon$.
Define $D_n$ as the set of linear combinations  with rational coefficients of characteristic functions of elements of $\mathcal A_n$. Since $\mathcal{A}_n$ is countable, so is $D_n$. Finally, define
$$D:=\bigcup_{N\geqslant 1}\left\{\sum_{i=1}^Nd_i,d_i\in D_i\right\}.$$
Then $D$ is countable and dense in $L^p(X)$.

How about simple function with rational coefficients?


Doubt in understanding polynomial theorem $m-n \mid p(m)-p(n)$ Consider a polynomial $P(x)=a_{n} x^{n}+\cdots+a_{1} x+a_{0}$ with integer coefficients.
The difference $P(x)-$ $P(y)$ can be written in the form:-
$
a_{n}\left(x^{n}-y^{n}\right)+\cdots+a_{2}\left(x^{2}-y^{2}\right)+a_{1}(x-y)
$
in which all summands are multiples of polynomial $x-y .$
Hence, If $P$ is a polynomial with integer coefficients, then $P(a)-P(b)$ is divisible by $a-b$ for any distinct integers a and $b$
My doubt is that why this is valid for integer coefficients, why we cannot replace integer with rational or real coefficients ???
I know that $a$ is divisible by $b$ if $a=bc$ where $$c is integer but here we are talking about polynomials and I read that in polynomials $P$ is said to be divisible by $Q$ if $P=QB$ where $B$ is another polynomial.

You have to differ between divisibility of polynomials and divisibility in integers.
It is true that $x-y\mid p(x)-p(y)$ but this is not necessary true in integers. Example $$x-\sqrt{2}\mid x^2-2$$ but this is clearly not true in integers, say if $x=1$ then $1-\sqrt{2}$ does not divide $-1$ since it is not defined in $\mathbb{Z}$.


Lets try plotting the following numbers on a number line: $-3, 0,e,\pi,-0.5+1.708i$.

It's clear that $-3\lt0\lt e\lt \pi$, but what to do with $-0.5+1.708i$ isn't so obvious.

Of course the reason for my confusion is that Complex numbers are not ordered. For, let's assume the opposite, that a complex number (e.g. $i$) is either zero, positive, or negative.
We know that $i\ne0$ because $0^2=0$ and $i^2=-1$. We know that $i$ is not negative because a negative time a negative is a positive and $i^2=-1<0$. Additionally we know that $i$ is not positive because a positive times a positive is a positive and $i^2=-1\lt 0$, thus contradicting our assumption that $i$ is either zero, positive, or negative. I know that this isn't a formal proof, but it does indicate that there are difficulties in ordering complex numbers.

This doesn't mean that Complex numbers don't have magnitude though. It should be obvious that $2i$ is twice as massive as $i$ and that $-1.0+3.416i$ is twice as massive as $-0.5+1.708i$. For more on that, see http://en.wikipedia.org/wiki/Complex_number#Absolute_value_and_argument

You have found that the polynomial $y=-6x^2 - 6x-19=0$ has no real roots, hence it cannot pass through the $y=0$ axis, so for all values of $x$ it always has the same sign, which indeed is (taking for example $x=0$)

\begin{equation*}
-6x^2-6x-19 = -(6x^2 + 6x + 19)  = -19< 0
\end{equation*}

hence, for all values of $x$, it is always a negative quantity (as said, it does not cross the $y=0$ line), so the inequality does not hold for any $x$ real.


Prove using natural deduction $((P \land Q) \rightarrow R) \vdash (P \rightarrow R) \lor (Q\rightarrow R)$ How do I prove the following using Natural Deduction?

$$((P \land Q) \rightarrow R) \vdash (P \rightarrow R) \lor (Q\rightarrow R)$$

My current approach:

So instead of proving $(P \rightarrow R) \lor (Q\rightarrow R)$, I figure I just need to prove $(P \rightarrow R)$ and use disjunction introduction.

The problem is my hypothesis requires $P$ and $Q$. How do I "create" this Q? I tried using the Law of the Excluded middle, but it does not seem to work.

Any help or insights is deeply appreciated.

This isn't quick using just the standard introduction and elimination rules. But the obvious brute-force way has the following shape using a Fitch-style lay out:

$((P \land Q) \to R)\\
\quad\quad |\quad \neg ((P \to R) \lor (Q \to R))\\
\quad\quad |\quad\quad|\quad (P \to R)\\
\quad\quad |\quad\quad|\quad ((P \to R) \lor (Q \to R))\\
\quad\quad |\quad\quad|\quad \bot\\
\quad\quad |\quad \neg(P \to R)\\
\quad\quad |\quad \vdots\\
\quad\quad |\quad P\\
\quad\quad |\quad \neg R\\
\quad\quad |\quad\quad|\quad (Q \to R)\\
\quad\quad |\quad\quad|\quad ((P \to R) \lor (Q \to R))\\
\quad\quad |\quad\quad|\quad \bot\\
\quad\quad |\quad \neg(Q \to R)\\
\quad\quad |\quad \vdots\\
\quad\quad |\quad Q\\
\quad\quad |\quad \neg R\\
\quad\quad |\quad (P \land Q)\\
\quad\quad |\quad R\\
\quad\quad |\quad \bot\\
\neg\neg((P \to R) \lor (Q \to R))\\
((P \to R) \lor (Q \to R))$

We temporarily assume the negation of the conclusion and aim for a reductio (what else?). Check that you understand the rationale at each step -- though the moves after that initial assumption are pretty much then the only sensible things to do. The dots hold the places for two routine proofs you need to know how to do (from something of the form $\neg(A \to B)$ to both $A$ and $\neg B$).

The OP ran into problems trying to use the Law of the Excluded Middle to prove this.  Here is a proof that uses the Law of the Excluded Middle (LEM):

The proof checker and associated text for more information are linked below.

Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/

P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/


Truth or lie? "At least one [of the two] of us is a liar." This problem involves two people. Person A and person B. They can
  either always tell the truth, or always lie. When asked, person A
  replies that: "At least one of us is a liar." Does person A and B
  tell the truth or do they lie?

What I have so far is that person A can either tell the truth or lie. If he tells the truth, then his statement can be expressed by "A$\lor$B", if he lies then it must be given that they are both speaking the truth, which contradicts that he would be a liar. So person A must be speaking the truth.

But if person A is speaking the truth, then person B could be either lying or speaking the truth. Both would still make person A tell the truth.

So how can I figure out whether person B is speaking the truth or not?

$$\begin{array}{cccc}
&&&\text{A says}\\
\text{A lies} &\text{B lies}&\text{A or B lies}&\text{A or B lies}\\
a&b&a\lor b&(a\lor b)\oplus a\\
F& F& F& \color{red}F\\
T& F& T& \color{red}F\\
F& T& T& \color{green}T\\
T& T& T& \color{red}F\\
\end{array}$$

I find it easier to always suppose that A, B tell the truth in any of these riddles with two types of people (that either always tells truth, or always lies). This is my approach:

$a$: A tells the truth

$b$: B tells the truth

Proposition "at least one of us is a liar" is $\lnot a \lor \lnot b$. Person A tells us this proposition, so $a$ and $\lnot a \lor \lnot b$ must have the same truth value (if A tells the truth, his proposition is true and if he lies, his proposition is false). Therefore, the answer to the problem is if there is one or more $T$ values of the proposition $a\leftrightarrow (\lnot a \lor \lnot b$). Let's see:

\begin{array}{cccccc}
a& b& \color{grey}{\lnot a}& \color{grey}{\lnot b}& \lnot a \lor \lnot b& a\leftrightarrow (\lnot a \lor \lnot b)\\
F& F& \color{grey}T& \color{grey}T& T& \color{red}F\\
T& F& \color{grey}F& \color{grey}T& T& \color{green}T\\
F& T& \color{grey}T& \color{grey}F& T& \color{red}F\\
T& T& \color{grey}F& \color{grey}F& F& \color{red}F\\
\end{array}

So, we see that $a\leftrightarrow (\lnot a \lor \lnot b)$ is true only when A tells the truth and B tells lies.

The same approach can be applied to any of this kind of problems.


Is $\mathbb{Z}\oplus\mathbb{Z}$ cyclic? I know there are infinitely many proper subgroups of $\mathbb{Z} \oplus \mathbb{Z}$, but is $\mathbb{Z} \oplus \mathbb{Z}$ cyclic?

Take a and b different integers.

Easy way to think is

If there is any generator, it can be either of the form
1)(a,a) or 2)(a,b)

If it is of form 1, you can never obtain element of second form.

Can you finish the argument?

Let's use @Arturo Magidin's hint:  $\Bbb Z\oplus\Bbb Z$ has the nontrivial subgroup $\Bbb Z\oplus\{e\}$, which is not of finite index.

$\therefore \Bbb Z\oplus\Bbb Z$ is not cyclic.

Reasoning behind the hint:   an infinite cyclic group is naturally isomorphic to $\Bbb Z$, whose only subgroups are of the form $n\Bbb Z$.


Intersection and containment detection for convex polygons I need to check if two convex polygons intersect of not, in a efficient way. The description of the intersection itself is not required.
I know the method of separating axis, but it takes $NM$ tests for polygons of sizes $N$ and $M$.
I also know the efficient methods based on rotating calipers (Toussaint), "edge chasing" (O'Rourke), sweepline (Shamos & Hoey), Voronoi diagrams (Yang & al.), which all take $N+M$ tests.
Unfortunately, all these are designed with the idea that the polygons do intersect and the focus is on computing the intersection. For this reason, they don't really describe what happens when the polygon outlines in fact do not intersect and how to cope.
I also know about the optimal approach from "Detection is easier than computation" (Chazelle), which takes $Log(N+M)$ tests but it is too sophisticated and complicated for my purposes.
So what I am after is a practical procedure not taking more than $N+M$ tests and able to tell if the polygons intersect of not, or if one is wholly inside the other.

I have found a suitable answer: this can be done by an extension of rotating calipers, using the separating axis principle and looking at the polygons as intersections of half-spaces.


Convergence of the infinite product of $\cos(1/\sqrt{i})$ I am trying to prove that the following product is convergent and show what it converges to:

$$\prod_{i=1}^\infty \cos{ \left( \frac{1}{\sqrt{i}} \right)}$$

I have heard that products are convergent if and only if the series of the $\ln$ of the product is convergent:

$$\ln \left[ \prod_{i=1}^\infty \cos{ \left( \frac{1}{\sqrt{i}} \right)} \right] = \sum_{i=1}^\infty \ln \left[ \cos{ \left( \frac{1}{\sqrt{i}} \right)} \right]$$

I have tried to prove that this is convergent in a number of ways to no avail. I believe it must therefore be divergent, but I have been having equally as much trouble proving this. It seems I have to prove that it is divergent using the comparison test; however, I was under the impression that this test only holds for series of positive reals.

In a neighbourhood of the origin,
$$ \log\cos x = -\frac{x^2}{2}-\frac{x^4}{12}-O(x^6) \tag{1}$$
hence
$$ \sum_{k=1}^{N}\log\cos\frac{1}{\sqrt{k}} = -\frac{H_N}{2}+O(1) = -\log\sqrt{N}+O(1)\tag{2} $$
and by exponentiating back we get that
$$ \prod_{k=1}^{N}\cos\left(\frac{1}{\sqrt{k}}\right) \approx \frac{C}{\sqrt{N}}\tag{3} $$
so the value of the given infinite product is just zero.

$\sum_{i=1}^\infty \ln(\cos{(\frac{1}{\sqrt{i}})})$

$\ln(\cos{(\frac{1}{\sqrt{i}})}) = \frac 12\ln((1-\sin^2{(\frac{1}{\sqrt{i}})})$

$\ln((1-\sin^2{(\frac{1}{\sqrt{i}})})<-\frac 1i$

$\sum_{i=1}^\infty \ln(\cos{(\frac{1}{\sqrt{i}})}) < -\frac 12\sum_{i=1}^\infty \frac{1}{i}$

$\sum_{i=1}^\infty \frac{1}{i}$ diverges so $\sum_{i=1}^\infty \ln(\cos{(\frac{1}{\sqrt{i}})})$ diverges.

$\lim_\limits{n\to\infty}  \sum_\limits{i=1}^n \ln(\cos(\frac{1}{\sqrt i})) = -\infty\\\lim_\limits{n\to\infty}  \exp(\sum_\limits{i=1}^n \ln(\cos{(\frac{1}{\sqrt{i}})})) = 0$


Solving $\frac{b}{a-b}e^{at}=\frac{x(t)}{a-x(t)}$ for $x(t)$ I`ve been trying to solve the differential equation $x(t)'=x(t)(a-x(t)), x(0)=b, t\in [0, \infty]$. Using the technique of seperation of variables, I get $\frac{b}{a-b}e^{at}=\frac{x(t)}{a-x(t)}$. Now I want to solve this for x(t). Any hints?

Why not to use simple method to find $x(t)$ by solving linear equation as follows $$\frac{a}{a-b}e^{at}=\frac{x(t)}{a-x(t)}$$ $$\implies a^2e^{at}-ae^{at}x(t)=(a-b)x(t)$$ $$\implies (a-b)x(t)+ae^{at}x(t)=a^2e^{at}$$ $$\implies \color{blue}{x(t)=\frac{a^2e^{at}}{ae^{at}+a-b}}$$ Now, for the given condition $x(0)=b$, substituting $t=0$, we have $$x(0)=\frac{a^2e^{a(0)}}{ae^{a(0)}+a-b}$$ $$=\frac{a^2}{a+a-b}=\frac{a^2}{2a-b}$$ $$\implies x(0)=b=\frac{a^2}{2a-b}$$ $$ab-b^2=a^2$$ $$a^2+b^2-2ab=0\implies (a-b)^2=0\implies \color{blue}{a=b}$$

since $\dfrac{1}{x(a-x)} = \dfrac{1}{a} (\dfrac{1}{x}+\dfrac{1}{a-x}) $ you have $$ \int (\frac{1}{x}+\frac{1}{a-x} )\mathrm{d}x = a \int \mathrm{d}t $$
so $$ Ln(x)-Ln(a-x) = a t + C $$
or
$$ Ln(\frac{x}{a-x})=at+C $$
so
$$ x(t) = \frac{a e^C e^{at}}{1+ e^C e^{at} } $$

now $x(0)=b$ so $b=\dfrac{a e ^c}{1+e^c} $ and $ C$ has been determined.


Proof that $(3^n - 2^n) / n$ is not an integer, $n \geq 2$ Trying to prove that $(3^n - 2^n)/n$ is not an integer for $n\geq 2$.

Was trying something along the lines of induction with:

$3^{n+1} - 2^{n+1} = 2(3^n - 2^n) + 3^n \equiv 0 \mod (n+1)$

But that gets messy...

Let $p$ be the smallest prime factor of $n$ and write $n = p^k m$ where $p \nmid m$. By repeated application of Fermat's little theorem it follows that $3^n - 2^n \equiv 3^m - 2^m \equiv 0 \bmod p$. If $p = 2, 3$ then this is clearly never possible; otherwise, it follows that $\left( \frac{2}{3} \right)^m \equiv 1 \bmod p$, hence that $\gcd(p-1, m) > 1$. But this is also impossible since the prime factors of $m$ are all larger than $p$.

I would consider first the case that $n \geq 3$ is prime.  Using Fermat's Little Theorem, you should be able to show $3^{n} - 2^{n} \equiv 1 \text{ (mod }n)$.

EDIT:  I retract what I said about the composite case.  I'll keep thinking about it.


undergraduate math vs graduate math It's really a mild, soft question.

So far, I am an undergrad student, contemplating on several majors.

What will be the major difference if I become a math major and go to a graduate school to study math?

Will a graduate math major generally focus on reading/studying recent research papers? Or will  the graduate math major be taught by professors on the things not covered in undergrad programs?

The reason why I am asking this question is that undergrad math programs seem to me at this point so well-covered that after studying the programs, a person will be able to do anything he wants to do, and if one wants to do research, he will be able to catch up with recent progress by reading research papers.

If this is true, why is the graduate school even needed?

By the way, I am a first-year undergraduate student :)

Here is what I tell my grad students:

The difference between undergrad mathematics and graduate mathematics is the difference between art history, or art appreciation, and learning to be an artist.

As an undergraduate you see a lot of mathematics, but you don't create new mathematics.  The goal of graduate school (and here I am speaking from experience with top fifty U.S. graduate schools, so what I am saying probably applies best in that context) is to learn how to create new mathematics, and then to create that new mathematics.

One specific consequence of this (in my view) is the following: often in undergraduate mathematics classes, proofs and rigor are presented almost as moral imperatives --- as if it is a moral failing to know a statement without knowing why it is true; consequently, people often put a lot of effort into learning arguments just for the sake of having learnt them.   (This is exaggerated, perhaps, but I think it reflects something real.)  On the other hand, in research, one learns arguments for different reasons: to learn technique, to pick out important ideas --- there is a professional aspect to the way one looks at pieces of mathematics which is not usually present in undergraduate mathematics.  One gives proofs in order to be sure that one hasn't blundered; one's interaction with the mathematics and the arguments is much more visceral than in undergraduate courses.

(I am not speaking from any experience now, but I think of the difference between learning how to interact with a block of marble, and bring a new form out of it, however rough it might be, in comparison to looking and learning about a lot of existing beautiful statues, masterpieces that they are.)

User2346:

Here's the opinion of someone who majored in math (at a top school in the United States), but was not a genius at it.

This is a condensed, generalized, down-and-dirty opinion, but I stand by it.

Your assumption about being well-covered is more accurate that not. If your intent is to form a mathematical basis of strength that you can apply to applications of math (e.g. finance stat econ) then undergrad is as far as you need to go with it. By the time you take Real Analysis & Complex analysis your mind will have developed in an amazing way. As Matt Pressland said, you learn how to learn Mathematics. Matt also mention the difference between that style of meta-cognition and being able to research. It's impossible to say today that you could or couldn't come up with a unique research idea that would be considered a contribution and that you're passionate about to pursue. However, unless the word research makes your heart throb, you don't need Math graduate school.

I've rewritten the rest of my answer to be a bit more compact.

Do not major in math, minor in it. You will gain a superior analytic mind. However, your employment potential depends on not being a Math major. The Corporate world (Actuarial aside) doesn't know wtf Math is. They will ask if you wanted to Teach High School.
Accounting is boring, and Economics is fluffy. Choose Statistics and augment with Finance. Also, Computer Science requires more hours spent at 1-sitting than Mathematics. Engineering trumps Computer Science as you can learn to be a programmer in your spare time.
First know what you want to do as a career. Do not decide this on your own. Go to your career center for expert help, and shadow professionals and intern. Funny how this is #3.

Conclusion
Math can be a minor. Let your possible career path determine your major. At the least you can be earning a salary when you're 22-23 while deciding what new career you want. At the best, you'll save yourself 3-5 years of searching, finding-professional-self, and rebuilding qualifications.

-

I decided to add some external links that would shed light on why some of my views are as they are. These come fro the Bureau of Labor Statistics... Oh look Statistics is even in the name =)

A small summary is included with the link, but please click and compare.
These are just samples, you can find your own BLS data to refute if you want.

Occupations from mostly-pure discipline

1. Economist. $43/hr. 6% projected growth. 15.4K jobs. Bachelors Entry-level
http://www.bls.gov/ooh/Life-Physical-and-Social-Science/Economists.htm

2. Mathematician. $48/hr. 16% projected growth. 3.1K jobs. Masters Entry-level
http://www.bls.gov/ooh/math/mathematicians.htm

3. Statistician. $35/hr. 14% projected growth. 21.5K jobs. Masters Entry-level
http://www.bls.gov/ooh/Math/Statisticians.htm

Occupations heavy in STAT with Finance Augment

4. Financial Analyst. $35/hr. 26% projected growth. 236.0K jobs. Bachelors Entry-level
http://www.bls.gov/ooh/business-and-financial/financial-analysts.htm

5. Actuary. $42/hr. 27% projected growth. 21.7k jobs. Bachelors Entry-level
http://www.bls.gov/ooh/math/actuaries.htm

Conclusion for the sample

Stat has way more job positions filled/needed much faster growth than Econ, same entry-level as Math.

Finance augmented,  Statistically heavy jobs are quite abundant, pay well, and grow fast. This is what I recommended. I recommended majoring in Stat with a Finance augment. Both examples of this require only a bachelors and are in a job space projected to grow much more than the Mathematician or the Economist. The pure Statistician indeed makes less than those two. So to repeat from previously: Math Masters not needed if you like other things. Don't major in Math in undergrad.


A problem on a sequence I have this question:

  Let $1,2,3,4,5,6,7,8,9,11,12,...$ be the sequence of all the positive integers which do not contain the digit zero. Write $\{a_n\}$ for this sequence. By comparing with a geometric series, show that $\sum_n \frac {1}{a_n} \lt 90$.

For the first part, I have found out the following steps to find $a_n$ when $n$ is given :

$1.$ First we see if there exists some positive integer $k$ such that $n= \frac {9^{k}-9}{8}$. If such a $k$ exists, then
$$a_n = 99 \cdots 99$$ where there are $k-1$ nines.

$2.$ If not, then we compute the integer $m= \lfloor log_9 (8n+9)\rfloor$ and write the integer $n- \frac {9^{m} - 9}{8}$ in nonary system. Suppose the representation looks like $(b_m b_{m-1} \cdots b_2 b_1 )_9$. If $b_r$ is the last non-zero digit, then
$$a_n = (b_m +1)(b_{m-1} +1) \cdots (b_{r+1} +1)(b_r)99 \cdots 99$$ where $(b_m +1)$ and $(b_{m-1} +1)$ and the similar terms are understood to be digits and there are $r-1$ nines.
Now my questions are:
$1.$ What does "Write $\{a_n\}$ for this sequence" mean?
$2.$ Is there any better method to get $a_n$? Is there a closed form for $\{a_n\}$?
$3.$ How to solve the second part of the question?

There are precisely $9^d$ $d$-digit numbers which do not contain the digit $0$. Each such $d$-digit number is clearly greater than $10^{d-1}$. Hence we have: $$\sum_{n\in\mathbb{N}}\frac{1}{a_n}<\sum_{d\in\mathbb{N}}\frac{9^d}{10^{d-1}} = 90$$

In fact, we always have that

$$\sum\limits_{n \in \mathbb{N}} \frac{1}{a_n} =(\sum\limits_{n \in \mathbb{N}} \frac{1}{n}) - (\sum\limits_{n \in \mathbb{N}} {\frac{1}{10 . n}}) = \frac{9}{10} (\sum\limits_{n \in \mathbb{N}} \frac{1}{n}).$$


If $L(x_1)$ = $L(x_2)$, does $x_1$ = $x_2$? Reviewing for an exam, and the book says False. But I can't seem think of a way to prove this wrong...

  If $L(x_1)$ = $L(x_2)$, does $x_1$ = $x_2$?

where $L$ is a linear transformation from $\Bbb R^n$ to $\Bbb R^n$, and $x_1$ and $x_2$ are vectors.

A trivial example would be the zero transformation
$$L: \mathbb{R}^n \to \mathbb{R}^n, L(x) = 0 \; \forall x \in \mathbb{R}^n.$$
Then $e_1 \neq e_2$ but $L(e_1) = 0 = L(e_2)$.

In general, not only in the context of vector spaces, $L(x_1) = L(x_2) \implies x_1 = x_2$ if and only if $L$ is injective. If, in particular, $L$ is a linear operator on vector spaces, then $L(x_1) = L(x_2) \implies x_1 = x_2$ if and only if $\text{ker}(L) = \{0\}$.

Consider x1 = -x2, and and L transforms the elements in a vector into absolute value. Then L(x1) = L(x2) even though x1 is not equal to x2.


A lower bound for de Polignac's formula De Polignac's Formula has many uses, for example when calculating the number of trailing zeroes of $n!$ :$$\nu_5(n!)=\sum_{i\le\lfloor\log_5n\rfloor}\left\lfloor\frac n{5^i}\right\rfloor.$$ For the purposes of my research, I'd like to find lower and upper bounds for $\nu_5(n!)$ that do not involve the floor or logarithm functions. I have found an upper one using the identity that $\lfloor\cdot\rfloor\le\cdot$ so that $$\nu_5(n!)\le n\sum_{i\le\lfloor\log_5n\rfloor}\frac1{5^i}=\frac n4\left(1-5^{\lfloor\log_5n\rfloor}\right)<\frac n4$$ by geometric series. The lower one causes problems since $$\nu_5(n!)\ge\sum_{i\le\lfloor\log_5n\rfloor}\left(\frac n{5^i}-1\right)=\frac{n}{4}\left(1-5^{-\lfloor\log_5n\rfloor}\right)-\lfloor\log_5n\rfloor\ge\frac{n-1}4-\log_5n$$ gives $$\frac{n-1}4-\frac{\ln n}{\ln5}\le\nu_5(n!)<\frac n4$$ and I cannot elementarily write $n$ in terms of $\nu_5(n!)$ only, for the lower bound.

  Does anyone have any further improvements on this, so that I can express $n$ as $f(\nu_5(n!))\le n<g(\nu_5(n!))$ for some functions $f$ and $g$?

Edit: The approximation $\ln n\approx an^{1/a}-a$ for large $a$ is not very useful since we would essentially be handling $a$th degree polynomials.

Well one lower bound is
$$\frac{n-4}{5}$$
 This works for all $n \ge 5$ as each time $n$ increases by $5$ an extra factor of $5$ is added to $n!$. For $n \in ${$9,14,19,24$} this lower bound is equal to the function.

You can solve $y = \frac{n-1}{4} - \frac{\ln n}{\ln 5}$ for $n$ using the Lambert W function:

$$ n = \frac{- 4 W_{-1}\left(- 5^{3/4 - y} \ln(5)/20\right)}{\ln(5)} $$


Two lines passing through $(-2,5)$ making angles $\arctan\frac 34$ & $\arctan\frac 23$ with the line $x-y+5=0$.Find the equations of the lines Two straight lines pass through the point $(-2,5)$ such that one of them makes an angle of $\arctan \frac 34$
with the given line $x-y+5=0$
and the given line makes an angle of $\arctan \frac 23$
with the other line . Find the equations of the two lines.
$\bf{Try} :$
Let the equation of the line be $y=mx+c$.$\cdots (1)$
If $(1)$ makes angle $\theta$ with the given line then $\tan\theta =| \frac {m-1}{1+m}|$
So by the given condition we have
$| \frac {m-1}{1+m}|=\frac 34$
Solving we get $m=7,\frac 17$
Putting these values of $m$ and using the condition that the line passes through $(-2,5)$ we get two lines as
$y=7x+19  \cdots (2)$
$7y=x+37\cdots (3)$
Similarly $| \frac {m-1}{1+m}|=\frac 23$
gives $m=5,\frac 15$ and we get the equations of the lines as
$y=5x+15\cdots (4)$
and $5y=x+27\cdots (5)$
But the answer is given as the lines $(2)$ and $(5)$. Why the lines $(3)$ and $(4)$ are excluded?

In coordinate geometry angles are usually measured in counter-clockwise direction from positive $x-$ axis. Your book is probably following that convention. You need to find two lines $L_1$ and $L_2$, where $L_1$ is the line that makes angle of $\arctan \frac34$ with the given line.
The given line makes an angle of $\frac \pi 4$ with positive $x-$ axis. So the angle that $L_1$ makes with positive $x-$ axis $=\frac \pi 4+\arctan \frac 34$ and so slope (gradient) of $L_1=\tan (\pi/4+\arctan \frac 34)=7$ and since this line passes through $(-2,5)$, its equation is: $y-5=7(x+2)$
Similarly, you may get equation of $L_2$.

{Red (2) Orange (3) Violet (4) Grey(5)} all for image if I will be able to upload image.
If question strictly wants positive angle for  statement then:
(3) makes $-\tan^{-1}(\frac {3}{4})$ with $x-y+5=0$
Similarly (4) makes $-\tan^{-1}(\frac{2}{3})$ with $x-y+5=0$
You may draw graph to check.


Is this vector parallel to the plane So I am given two points $A=(-.5,2.3,-7.3)$ and $B=(-2,17.1,-0.3)$ and then using $AB = OB - OA$ to give me $(1.5,-14.8,-7)$. The plane is $$x+23y+13z=500$$ From there I got $r.n$ where $r=(1.5,-14.8,-7)$ and $n=(1,28,13)$. From here I do not know how to check if the vector is perpendicular to the plane.

We have the two points $A(-0.5,2.3,-7.3)$ and $B(-2,17.1,-0.3)$, as well as the general equation of the plane $x+23y+13z-500=0$. The vector from $A$ to $B$ can easily be found.
$$\vec{\mathbf{r}}=\langle(-2+0.5),(17.1-2.3),(-0.3+7.3)\rangle=\langle-1.5,14.8,7\rangle$$
The normal vector to the plane is $\vec{\mathbf{n}}=\langle1,23,13\rangle$. These two vectors are parallel if their cross product is equal to $\vec{\mathbf{0}}$ [1]. If $\vec{\mathbf{r}}$ and $\vec{\mathbf{n}}$ are parallel vectors, then they are both perpendicular to the plane in question.
$$\vec{\mathbf{r}}\times\vec{\mathbf{n}}=\langle(14.8)(13)-(7)(23),(7)(1)-(-1.5)(13),(-1.5)(23)-(14.8)(1)\rangle$$
This comes from the fact that $\vec{\mathbf{u}}\times\vec{\mathbf{w}}=\langle u_2w_3-u_3w_2,u_3w_1-u_1w_3,u_1w_2-u_2w_1 \rangle$ [2].
$$\vec{\mathbf{r}}\times\vec{\mathbf{n}}=\langle31.4,26.5,-49.3\rangle\ne\langle0,0,0\rangle$$
Let us see if the vector $\vec{\mathbf{r}}$ happens to be parallel to the plane in question. If it is, then the dot product of $\vec{\mathbf{r}}$ and $\vec{\mathbf{n}}$ will be $0$ as the two vectors will be orthogonal.
$$\vec{\mathbf{r}}\cdot\vec{\mathbf{n}}=(-1.5)(1)+(14.8)(23)+(7)(13)=-1.5+340.4+91=429.9\ne0$$
The vector from $A$ to $B$ is not perpendicular to the plane in question, nor is it parallel to it.

In general two vectors: $(a_1, a_2, a_3)$ and $(b_1, b_2, b_3)$ are said to be parallel if
$$\frac{a_1}{b_1}=\frac{a_2}{b_2}=\frac{a_3}{b_3}$$
Now, the vector $\vec{OB}=(-1.5,14.8,7)$ will be perpendicular to the plane:$x+23y+13z=500$ iff  vector $r=(-1.5,14.8,7)$ is parallel to the normal vector $n=(1,28,13)$ of plane i.e.
$$\frac{-1.5}{1}\ne\frac{14.8}{28}\ne \frac{7}{13}$$
above inequality shows that vector $r=(-1.5,14.8,7)$ is not parallel to the normal vector $n=(1,28,13)$ i.e. vector $r=(-1.5,14.8,7)$ is not perpendicular to the given plane.


Walking the circle in steps We're walking on a circular path by making a step forward and turning right a constant angle each time.

If we shorten our step n times, how should the angle change, such that circular path we're walking remains the same?

I will assume the start and end points of each step lie on a fixed circle which has radius $R$. Then if each step has length $l$ and the angle turned between one step and the next is $\theta$, then $\theta$ is also the angle of the arc between the start and end points of each step. So

$l = 2 R \sin \left( \frac {\theta}{2} \right)$

If our step length changes to $l'$ then the angle turned between one step with length $l'$ and the next will now be

$\theta' = 2\sin^{-1} \left( \frac{l'}{2R} \right)$

At the point where we change our step length from $l$ to $l'$ then the angle between the last step with length $l$ and the first step with length $l'$ will be

$\frac{\theta + \theta'}{2}$

Well i guess you would be walking in a regular cyclic polygon. So if the number of sides(steps) is n, the angle will be $\frac{n-2}{n}\cdot\theta$
You will be able to calculate the change yourself.


Simplifying this boolean function How can I completely simplify this equation using algebraic simplification rules?

$$x'y'z + x'yz + xyz$$

We use the distributive law twice:

$$\begin{align} x'y'z + x'yz + xyz & = x'z(\underbrace{y' + y}_{= 1}) + xyz \\ \\ & = x'z + xyz \\ \\ &= z(x' + xy)\end{align}$$

What counts as simplified depends on the context.

We can further simplify by the distributive property:

$$z(x' + xy)=z((x'+x)(x'+y))=z(x'+y)$$
where $x'+x$ is always true.

Check out:
http://en.wikipedia.org/wiki/Boolean_algebra_(structure)#Definition


Solve for x in $3(1-e^{-x})=x$. $3(1-e^{-x})=x$ . We nees to find the value of $x$ from the equation.

I have tried using expansion of exp but thats not fruitful....any ideas are welcome...

Thank You...

$$x-3=-3e^{-x}$$
$$(x-3)e^x=-3$$
$$(x-3)e^{x-3}= -3e^{-3}$$
$\begin{cases} X=x-3 \\ Y=-3e^{-3} \end{cases} \quad\to\quad Xe^X=Y \quad\to\quad X=W(Y)$

$W$ is the Lambert's W function.
$$x-3=W\left(-3e^{-3}\right)$$
$$x=3+W\left(-3e^{-3}\right)$$
The Lambert's W function is a multi valuated function :

http://mathworld.wolfram.com/LambertW-Function.html

First real value : $W\left(-3e^{-3}\right)=-3$ and second real value : $W\left(-3e^{-3}\right)\simeq -0.17856$

The real solutions are : $x=0$ and $x\simeq 2.82144$

In practice, there is no need for Lambert-W function. One can directly compute an approximate of the root thanks to numerical methods, for example Newton-Raphson.

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

Let $y = 1-\frac {x}{3} \Rightarrow x=3(1-y)$

$e^{-3(1-y)}=y$

$e^3 \cdot e^{-3y}=\frac {1}{y}$

$e^{-3y}=\frac {e^{-3}}{y}$

Powering both sides by $\frac {e^{-3}}{y}$ to eliminate $y$ in the left side

$e^{-3e^{-3}}={\frac {e^{-3}}{y}}^{\frac {e^{-3}}{y}}$

Base and power are equal for both sides $a^a=z^z$, so one of the solutions is to let

$e^{-3}=\frac {e^{-3}}{y} \Rightarrow y=\frac {e^{-3}}{e^{-3}}=1$

then $x=3(1-y) =0$

Note that the super square root of $a^a$ may has 2 real values.

for example $z^z = {\frac {1}{2}}^{\frac {1}{2}}$ has 2 real solution
$z= \frac{1}{2}$ and $z= \frac {1}{4}$


Use implicit differentiation to find the largest y-value in the loop of the Folium of Descartes, which is given by :$ x^3 + y^3 - 3xy = 0$. My professor didn't even explain how to do this. I asked the TA,and they told me to equal the first derivative to $0$ and solve for $x$ and then see if the second derivative is less than $0$? Can someone explain this problem to me?

First, we do the implicit derivative to simplify our equation. Because maxima/minima occur when $f'(x)=0$, we take the implicit derivative and set it equal to 0. We should end up with a few maxima/minima. To tell whether if it is the maxima or minima, we use the second derivative test to see whether it is a maxima or minima(If $f''(x)<0$ then it is a maxima ).

Yep, that's exactly what you gotta do: ' equal the first derivative to 0 and solve for x and then see if the second derivative is less than 0'... Assuming you know implicit differentiation it should be easy - and in case you don't - you do know it, it's basically the chain rule, differentiate everything by x, applying chain rule to y.


Calculate asymptotic upper bound for a function consider the following function $f(n) =$  $n^{4.5}-$$(n-2)^{4.5}$
I want to calculate a good tighter upper bound for it.
I calculated a lower bound for it like -
$n^{4.5} - (n-2)^{4.5} >= n^{4.5} - \sqrt n * (n-2)^3$
Now if we expand $(n-2)^3$ the $n^{4.5}$ term will cancel out and we will have a $n^{3.5}$ term
So $f(n) \geq c * n^{3.5}$ for some constant $c$ we can say...
But getting a good upper bound seems confusing to me... Also not sure if this lower bound is a tight one.
Thanks


Abelian Group Element Orders I want to show that if a finite abelian group has elements of order $m$ and $n$ then it will have an element of order $\text{lcm}(m,n)$.

First I proved the lemma if $a$ has order $m$ and $b$ has order $n$ with $m,n$ coprime, then $ab$ has order $mn$. This is because the subgroups $\langle a \rangle$ and $\langle b \rangle$ have trivial intersection (because the order of their intersection must divide their orders $m$ and $n$, which is a consequence of it being a subgroup of both), which implies $a^{i} \in \langle b \rangle$ iff $i \equiv 0 \pmod n$, and similarly $b^{i} \in \langle a \rangle$ iff $i \equiv 0 \pmod m$. Using that we deduce that $(ab)^i = a^i b^i = 1$ iff $i \equiv 0 \pmod {mn}$.

So if $a$ was an element of order $m$ and $b$ and element of $n$ with $g = \gcd(m,n) \not = 1$ I thought that $\text{lcm}(m,n) = \frac{m}{g}n$ so and $\frac{m}{g},n$ are coprime so I should construct an element (from $a$) with order $\frac{m}{g}$ then conclude the theorem by the lemma. For the construction I think it's just $a^g$.

I just have a nagging doubt about the correctness of the second proof, did I miss some important detail? Also if there are any neater ways to prove this (which don't depend on the structure theorem) I would like to learn them too.

Saying that $m/g$ and $n$ are coprime is wrong. Consider $n=12$, $m=18$, the standard example. What is true is that $m/g$ and $n/g$ are coprime. But if you want to complete the proof using the elements $a^g,b^g$, then you will realise that although these now have coprime orders, the least common multiple (=product) of those orders is no longer $\def\lcm{\operatorname{lcm}}\lcm(n,m)$, but $\lcm(n,m)/g$.

So instead another idea is needed. The best I have been able to think of is treat all prime numbers $p$ dividing $nm$ separately; if $p$ divides the order of only one of $a,b$ then there is nothing to do for $p$, but if it divides both orders, then keep the element in which the multiplicity is highest (choose one in case of a tie) and kill the factors $p$ in the order of the other element (say it was $b$) by replacing $b$ by $b^{p^i}$ (where $i$ is the minority multiplicity). This modification is designed to leave the least common multiple of the orders of $a,b$ intact, and when all primes have been processed the greatest common divisor has become$~1$. After these modifications, the element $ab$ has the desired order.

Everything you said seems fine.  Using additive notation, you only need to consider the subgroup $sa + tb$ of $G$.  This group contains the element $a+b$ which has the property

$$ \hbox{lcm}(m,n) \cdot (a + b) = 0 $$

so the order of $a+b$ is at most $\hbox{lcm}(m,n)$.

Moreover the order must be equal to $\hbox{lcm}(m,n)$, because any smaller order could be divisible by both $m$ and $n$.


Finding the domain of $\log(\log(x))$ Why is the domain of $f(x)=\log(x)$ is $x>0$ but the domain of $f(x)=\log\log(x)$ is $x>1$.

Why is there a difference? (default value of base is 10)

If $f(x)=\log(\log x)$, then it has to be $x>0$, but the argument of the most external logarithm, argument which is $\log x$ also has to be greater than $0$.

So $\log x>0$ is equivalent to $x>10^0$, that is $x>1$. Then you need both $x>0$ and $x>1$ to be true, and you can just say that $x>1$. Then, $(1,+\infty)$ is the domain of $f$.

The domain of the function $g(x)=\log_{10}(f(x))$ is the set of all $x$ such that $f(x)>0$. In particular, if $f=\log_{10}$ then $$\log_{10}(x)>0 \quad \textrm{iff} \quad x>10^0 =1$$


Prove that the square root of any irrational number is irrational. The problem I'm having with this proof is that I'm not sure if my proof actually proves the theorem correct or if I'm using circular reasoning.

Theorem:

Prove that the square root of any irrational number is irrational.

Proof:

=> Suppose not. The square root of any irrational number is rational.

=> Let $m$ be some irrational number. It follows that $\sqrt{m}$ is rational.

=> By definition of a rational number, there are two positive integers $p$ and $q$ such that $\sqrt{m} = \dfrac{q}{p}$

=> $m = \dfrac{q^{2}}{p^{2}}$

=> $q^{2}$ and $p^{2}$ are integers, and by definition of a rational number, $\dfrac{q^{2}}{p^{2}}$ is rational

=> $m$ is irrational and is equal to the rational number $\dfrac{q^{2}}{p^{2}}$. This is a contradiction.

=> Thus, the square root of any irrational number is irrational.

I've seen this proof also done with the addition of these steps:

$m = \dfrac{q^{2}}{p^{2}}$

$m \times p^{2} = q^{2}$

Because a an irrational number times a rational number is irrational, we have an irrational number equaling a rational number which is a contradiction.

My question is: Is this step really necessary?

Said shortly, $$\left(\frac pq\right)^2=\frac{p^2}{q^2}$$ is rational.

The square of any rational is rational, hence no rational is the square root of an irrational.

Yes, because you are going round in circle's if you don't say the extra step. It is like proving that cats are better than dogs, and assuming 'if cats are not better than dogs' but this is wrong because 'blah blah blah' and then ending your proof. You have not said anything much, except a>b so b


Determinant with variables What is the following determinant?

$$\begin{vmatrix}1+a & b & c & d \\a & 1+b & c & d \\a & b & 1+c & d \\a & b & c & 1+d \end{vmatrix}$$

I calculated it as $0$ but I do not think it is right. Thanks in advance.

Subtract row $2$ from row $1$, row $3$ from row $2$, and row $4$ from row $3$.  You get
$$
\begin{bmatrix}
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 \\
a & b & c & 1+d
\end{bmatrix}
$$
If you expand along the fourth row, all the $3 \times 3$ determinants will be upper or lower triangular.  So, the determinants will be the product of the diagonals.

Let $\mathrm v := \begin{bmatrix} a & b & c & d\end{bmatrix}^{\top}$. Using the Weinstein-Aronszajn determinant identity,
$$\det \begin{bmatrix}1+a & b & c & d\\ a & 1+b & c & d\\ a & b & 1+c & d\\ a & b & c & 1+d\end{bmatrix} = \det \left( \mathrm I_4 + 1_4 \mathrm v^{\top}\right) = 1 + \mathrm v^{\top} 1_4 = \color{blue}{1 + a + b + c + d}$$


What are the common abbreviation for minimum in equations? I'm searching for some symbol representing minimum that is commonly used in math equations.

If you are looking for a symbol (and for some reason want to avoid the clear, and much more common notation "$\text{min}$"), I believe that some people use $\wedge$ -  see here.

The choice of symbol, depends mainly on the mathematics, physics, or programming discipline the equation is used in.

Merrian-Webster defines minimum as "the least quantity assignable, admissible, or possible."

For ordering, minimum means 'less than or equal to', which
is symbolized in some/many mathematics disciplines as ≤.
List A = [1 0 1 0], List B = [1 1 0 0];  List (A[i]≤B[j] = [1 0 0 0].
For Boolean algebras, just multiply the two lists to generate the minimums:
List (A[i]*[j] = [1 0 0 0].  For Boolean Algebra * is a 'minimum' operator,
whereas + is a 'maximum' operator.

But, as mentioned above,  'min' is OK, as most programming languages
use some variant of min(). but C does not. Program comments would use either
'min' or '<=, both easy ASCII typing.

Stochastic processes use ∧ for minimum; however mathematical logic uses ∧
to mean logical conjunction or AND.


Factoring Cubic Equations I’ve been trying to figure out how to factor cubic equations by studying a few worksheets online such as the one here and was wondering is there any generalized way of factoring these types of equations or do we just need to remember a bunch of different cases.

For example, how would you factor the following:

  $x^3 – 7x^2 +7x + 15$

I'm have trouble factoring equations such as these without obvious factors.

Your example has integer coefficients. We deal mainly with that case, or the essentially equivalent case of rational coefficients, since we can always multiply through by the lcm of the denominators and get integer coefficients,

The following result, called the Rational Roots Theorem, is sometimes useful.

Let $P(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots +a_0$ be a polynomial with integer coefficients $a_i$, and suppose $a_0\ne 0$. Then any rational root of $P(x)=0$  must have shape $\frac{c}{d}$, where $c$ divides $a_0$ and $d$  divides $a_n$,

In our case, $c$ must divide $15$ and $d$ must divide $1$. That gives the candidates $\pm 1$, $\pm 3$, $\pm 5$, and $\pm 15$.

Test them all. We get very lucky, $3$ works. To find the other roots, we can divide our polynomial by $x-3$, getting a quadratic, which we solve using the Quadratic Formula.

This procedure often works when we are dealing with a "made up" problem. In particular, if the cubic is in a school exercise, almost certainly it has a rational root.  But the procedure certainly does not always work. Most cubics with integer coefficients do not have a rational root.   And the procedure  only applies to polynomials with rational coefficients.

If there are no rational roots, we are in trouble. True, there is a formula for the roots of the general cubic, usually called the Cardano Formula. It dates back to the sixteenth century. One can also find formulas that use trigonometric or hyperbolic functions. But overall the formulas are not very useful.

There are good methods for quickly approximating the roots to high accuracy,


Solve $15\cdot x=27\pmod{18}$ EXERCISE:

  Solve   $15\cdot x\equiv 27\pmod{18}$

SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.

So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!

Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equiv 9\pmod{6}$

So, from now and then I can't understand how to proceed!

The book continue with this:
"We have a unique solution of   $5\cdot x\equiv 9\pmod{6}$ which is":

$x=5\cdot 9 =45\equiv 3\pmod 6$ and $15\cdot 3\equiv 27\pmod{18}\equiv 9\pmod{18} $

The other solutions are:$(9,15)$. So all solutions are these: $(3,9,15)$

Can anyone explain me how the author of the book continue the exercise?I really can't understand how he ends up with these solutions!Is there another way to reach these solutions?

I would really appreciate a thorough  explanation, since I've just started working on these type of exercises and I have to clear my mind on them.

Thanks for your time !

Note that

$$15\cdot x\equiv 27\pmod{18}\iff 15\cdot x\equiv 9\pmod{18}$$

and

$$15\cdot x\equiv 9\pmod{18}\iff 15\cdot x=9+18\cdot k\\\iff 5\cdot x=3+6\cdot k \iff 5\cdot x\equiv 3\pmod{6}$$

note also that

$$5\cdot 5\equiv 25 \equiv 1\pmod{6}$$

then

$$5\cdot x\equiv 3\pmod{6}\iff 5\cdot 5\cdot x\equiv 5\cdot3\pmod{6}\iff x\equiv 15\pmod{6}\\\iff x\equiv 3\pmod{6}$$

It is possible to solve this problem using algebraic symbols or augmented matrices. For example the following uses augmented matrices to solve the problem:

$$\begin{array}{c|c|c|}
 \hline
 \text{x} & \text{y} & \text{c} & \text{s} & \text{t} \\
\hline
\text{15} & \text{-18} & \text{27} \\
& \text{+3} & \text{+3} & \text{-15} \\
 & & & & \text{3} \\
\hline
\text{5} & \text{-6} & \text{9} \\
& \text{1} & \text{1} & \text{-5} \\
\hline
\text{1} & & \text{3} & \text{9} \\
& \text{1} & \text{1} & \text{-5}
\\
\hline
\end{array}$$

I have included animations for the construction of the augmented matrices in the PDF document "How I solved 15x≡27 (mod 18)" and a description of the "method" and its relation to the Euclidean Algorithm and the Extended Euclidean Algorithm.


Expressing in the form $A \sin(x + c)$ Express in the form $A\sin(x+c)$
  a) $\sin x+\sqrt3\cos x$; b) $\sin x-\cos x$

  sol: a) $A=\sqrt{1+3}=2$, $\tan c=\frac{\sqrt 3}1$, $c=\frac\pi3$. So $\sin x+\sqrt3\cos x=2\sin(x+\frac\pi3)$
  b) $\sqrt 2\sin(x-\frac\pi4)$

Can someone please explain the method used in the provided solution above? (I'm not familiar with this way of solving whatsoever.)

Thanks in advance =]

The anonymous commenter has already answered your question, but in case you have any remaining doubts, I will provide a detailed answer.

For starters, do note that

$A \, \sin (x + c) = \left(A \, \cos(c) \right) \, \sin(x)  + \left( A \, \sin(c)\right) \, \cos(x)$

Since you have $\sin (x) + \sqrt{3} \, \cos (x)$, it follows that $A \, \cos(c) = 1$ and $A \, \sin(c) = \sqrt{3}$. Therefore, since $\sin^2 (x) + \cos^2 (x) = 1$, we have that $A^2 = 4$, which yields $A = 2$, and $2 \cos (c) = 1$, which yields $c = \pi / 3$. Finally, we conclude that

$\sin (x) + \sqrt{3} \, \cos (x) = 2 \sin (x + \pi / 3)$.

I ran into the same problem with the MIT OCW problem set.  I found out that A represents the amplitude and c represents the phase shift.  I sketched the sine and cosine curve then the result of sin - cos.  Where sin = cos, the result is zero at pi/4 and 3pi/4, so that makes c= -pi/4.  the max amplitude is at 3pi/4 where sin = (square root of 2)/2 and cos = - (sqr2)/2, making sin-cos square root of 2, so that's the amplitude.  so the answer is sinx-cosx = sqrt(2)(x-pi/4)


$\forall x,y\in \mathbb N\ ,\ \exists\ z\in\mathbb N$ that $x+z$ is square but $y+z$ is not square. I am trying to prove this:

  $\forall x,y\in \mathbb N$ and $x\neq y,\  \exists\  z\in\mathbb N$ that $x+z$ is
  square but $y+z$ is not square.

$\mathbb N$ is set of natural numbers.
Can you please suggest a hint?

You know how far apart $x$ and $y$ are.  Look for squares that are further apart than that.

Suppose the contrary; that is, suppose that for each pair $x$, $y$ of distinct natural numbers, the following is true:

For every natural umber $z$, if $x+z = m^2$ for some natural number $m$, then $y+z = n^2$ for some (other) natural number $n$.

Obviously, $m \neq n$ since $x\neq y$.

Then we must have $$z = m^2 - x = n^2 - y,$$
and so we must have $$x-y = m^2 - n^2,$$
Which is clearly false, for example, when $x=2$ and $y=1$ for the smallest difference (in absolute value) between the squares of two natural numbers is $3$. In any case, there are no natural numbers $m$ and $n$ whose squares differ by $1$. So a contradiction arises, and our assertion is proved.


Prove that if $2^x,3^x, 5^x, 7^x, 11^x ... $ are all integers then $x$ is an integer as well How easy is it to prove that if $2^x,3^x, 5^x, 7^x, 11^x ... $ are integers then $x$ is an integer as well? I have read the definition of the exponent functions as given in my calculus text, and the question propped up.

Given $ 2^x $, $ 3^x $, $ 5^x $,... are integers. Now suppose $ x $ is not an integer. Suppose it is rational of the form $ p/q $ where $ gcd(p,q) = 1 $.
So $ 2^{p/q} $, $ 3^{p/q} $, ... can't be integer, since base of exponents are all prime.


If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root then which of the following is true If the equation $x^2+px+q=0$ and $x^2+qx+p=0$ have a common root then which of the following is true:

$1$. $p=q$

$2$. $p+q+1=0$

$3$. $p+q=0$

$4$. $\textrm {both} 1,2$.

MY WORK:

$$x=\dfrac {-p\pm \sqrt {p^2-4q}}{2}$$
And,
$$x=\dfrac {-q\pm \sqrt {q^2-4p}}{2}$$

HOW DO I GET IT COMPLETED?

If $$x^2+px+q = 0 = x^2+qx+p$$ then $$px+q=qx+p$$ So either $p=q$ or $x=1$ and thus $1+p+q=0$.

So one of (1) or (2) must be true (it is not necessary for both to be true.) It's not clear what "both" is doing in the question


Bridge hand Combination/Permutation A Bridge hand consists of 13 cards from a deck of 52 cards. In how many ways can a (bridge) hand
consisting of 5 spades(♠), 4 hearts(♥), 4 diamonds(♦) and 0 clubs(♣) be selected?

There are $13 \choose 5$ ways of choosing $5$ spades.

There are $13 \choose 4$ ways of choosing $4$ hearts.

There are $13 \choose 4$ ways of choosing $4$ diamonds.

So there are $13 \choose 5$$\times$$13 \choose 4$$\times$$13 \choose 4$ ways of selecting a bridge hand with $5$ spades, $4$ hearts, $4$ diamonds, and $0$ clubs.

A slightly different answer is also possible, depending on whether the question is "how many ways can a (bridge) hand ... be selected", or whether it means how many possible different hands there are. The previous answer is how many possible different hands there are. But how many ways you can select those hands would be 13P5 * 13P4 * 13P4  (not 13C5 * 13C4 * 13C4).
Edit 20160416: Thanks for the comment Thomas - silly mistake by me. It is indeed (13C5*13C4*13C4)*13C1 as you noted.


Inverse of function, containing a fraction This is basic, I know, but I cannot seem to come up with the right answer.

Find the inverse of the function:
$$f(x)= \frac3{x+1}$$

My steps:
1. Convert f(x) to y
$$y = \frac3{x+1}$$

Switch places of x and y
$$x= \frac3{y+1}$$
Try to solve for y. So I multiply the denominator by x to get rid of it
$$x(y+1) = 3$$
After multiplying, I'm left with
$$xy + x = 3$$
Which then converts to
$$2xy = 3$$
Then I get rid of 2x on the left, placing it on the right
$$y = 3 - 2x$$
Now I convert y to the inverse function
$$f^{-1}(x) = 3 - 2x$$

My answer is obviously wrong. The correct answer is:
$$f^{-1}(x) = \frac{3-x}{x}$$

Where did I mess up?

Thanks!

Step 4 is wrong, check it. instead, divide by x after step 2, and carry over 1.

$XY+X=3
\\XY=3-X
\\Y=(3-X)/X$

This is correct steps after step $3$


Is it true that $a$ can't be zero in the quadratic function $y=ax^2+bx+c$? I read that for $y=ax^2+bx+c$ is a quadratic function where $a\neq0$, but is it true that $a$ really can't be zero?  I think it is because if $a$ was zero, there wouldn't be a parabola.  There would just be a flat line, so then it wouldn't be quadratic because the $x^2$-term indicates if the parabola opens upward or downward.  Is this right or is it true about what I asked?

If $a=0$, you no longer have a parabola.

Instead, you have a line: $y = bx+c$, with slope equal to $b$, and a $y$-intercept at $c$.

I guess I'm right.  If $a$ is zero, then $y=ax^2+bx+c$ would change to $y=bx+c$, leaving $b$ as the slope and $c$ as the y-intercept, leaving a flat line, not a parabola.


How many positive integers are factors of a given number? I've been trying to find / generate a formula for the following problem:

Given a number, how many positive integers are factors of this number.

In practice, you could simply build a table as such (lets assume the number is 36):

1 |  36
2 |  18
3 |  12
4 |  9
6 |  6

Thus there are 9 positive integers that are factors to 36.

This method seems like it would be taxing if the number was large. So I was thinking that if you knew the prime factorization for a number such as  $\,2 \cdot 2 \cdot 3 \cdot 3 = 36, \,$ there should be a formula for the number of unique combinations you can multiply these prime factors together. That's where I am stuck. Thanks in advance!

Hint: if the prime factorization of $\,n\in\Bbb N\,$ is

$$n=\prod_{k=1}^rp_k^{a_k}$$

then the number of divisors of $\,n\,$ is

$$\prod_{k=1}^r(a_k+1)$$

Let me explain this with the help of an example.
Let the number be 630.
Now the prime factorization of $630 = 2^1 × 3^2 × 5^1 × 7^1$.
Any factor is a combination of one or more numbers with their power ranging from 0 to maximum.

For 2 , the power can be 0 or 1 i.e. 2 ways.
For 3 , the power can be 0 or 1 or 2 i.e. 3 ways.
For 5, the power can be 0 or 1 i.e. 2 ways
For 7 , same as 5 i.e. 2 ways.

Now if you know basics of permutation and combination, the number of ways we can select any power of the above prime factorization is

2×3×2×2 = 24.

Hence 630 has 24 factors.

Now the formula is Let  $N = (A^r)×(B^m)×(C^n)$ where A,B,C are the prime factors of N, then the total number of positive factors of N is given by (r+1)(m+1)(n+1).


Complex numbers simplify I'm new here and I'm studying complex numbers is there a way the simplify this:

$$\left(\frac{45\sqrt{3}i}{i^\pi}\right)^{\pi \times 5}$$


Second Order Eigenvector In my differential equations class, my professor calls solutions of the following matrix equation second order eigenvectors:

$$(A-LI)^2 v = 0.$$

A vector $v$ that satisfies this equation where $L$ is an eigenvalue, $A$ is a matrix and $I$ is the identity matrix is called a second order eigenvector.

I can't find any information on this topic so I suspect it goes by a different name. If so, what is the name?

Also in particular, I need to know how to find second order eigenvectors.

Thanks.

P.S.: To add context, we are solving linear differential equations of the form $v' = Av$.
The particular case we are looking at is when we have a double real eigenvalue and only one eigenvector.

My professor said the solution is of this form where $L$ is the eigenvalue, $w$ is a eigenvector, and $u$ is a second order eigenvector:

$$e^{Lt}w + te^{Lt}u.$$

Here is an example:
$$A= \begin{bmatrix} 1 & -1 & 1 \\ 0 & 0 & 1 \\ 0 & -1 & 2 \end{bmatrix}.$$

After finding $L=1$ to be the only eigenvalue, I get this:
$$A - LI = \begin{bmatrix} 0 & -1 & 1 \\ 0 & -1 & 1 \\ 0 & -1 & 1 \end{bmatrix}.$$

If I square this though to solve $(A-LI)^2 v = 0$, I get the zero matrix, which means I don't get a unique second order eigenvector.


Linear combination of two vectors that zeroes out the quadratic form of matrix. Let $A$ be a given $N\times N$ hermitian matrix. Let $v$ be a vector such that $v^HAv>0$ and $w$ be a vector such that $w^HAw<0$ ($w^H$ denotes conjugate transpose of $w$). Assume both $v$ and $w$ are unit-norm. Does there exist any non-zero vector $x$ which is a linear combination of $v$ and $w$, such that $x^HAx=0$?

A non-constructive proof:

Let $Q(\mathbf u)=\mathbf u^HA\mathbf u$. Since $Q(k\mathbf u)=\left|k\right|^2Q(\mathbf u)$, we know that $\mathbf w$ is not a scalar multiple of $\mathbf v$. Consider a parametrization $\mathbf c(t)$ ($t\in\mathbb R$) of the unit circle in the plane spanned by $\mathbf v$ and $\mathbf w$ with $\mathbf c(a)=\mathbf v$ and $\mathbf c(b)=\mathbf w$. $Q(\mathbf c(t))$ is a continuous function of $t$, so there must be some $t_0\in(a,b)$ for which $Q(\mathbf c(t_0))=0$.

A constructive proof:

$\exists \lambda = -\frac{v^HAv}{w^HAw} \in \mathbb{R}^*_+$ such as $v^HAv+\lambda w^HAw = 0$.

if $w^H Av =0$ then we can construct $u=v+\lambda^{1/2}w$ which verify:

$u^HAu=v^HAv+\lambda w^HAw + \lambda^{1/2}(w^H Av+v^H Aw)=0$

else let $\tilde w = i(w^HAv) w$ then $ \tilde w^H A \tilde w  = (-i)(i)|w^H Av|^2 w^H Aw < 0$

so there exists $v$ and $\tilde w$ and $\tilde \lambda= -\frac{v^HAv}{\tilde w^HA\tilde w}  \in \mathbb{R}^*_+$ such as $v^HAv+\lambda \tilde w^HA\tilde w =0$ in one hand.

In other hand  $\tilde w^HAv = (-i) (\overline{w^HAv}) (w^HAv) =- \overline{\tilde w^HAv}$.

So $(v+{\lambda}^{1/2}\tilde w)^HA(v+{\lambda}^{1/2}\tilde w) = v^HAv+\tilde \lambda \tilde w^HA\tilde w + \tilde \lambda^{1/2}(\tilde w^HAv + \overline{\tilde w^HAv}) = 0$


Expression as a product of disjoint cycles Let $\alpha = (9312)(496)(37215) \in S_n, n \ge 9$. Express $\alpha$ as a product of disjoint cycles.

I know this is probably a really easy question, but my professor didn't elaborate on how to exactly do this and neither does my assigned text. If anyone could elaborate on the algorithm of going about this I would really appreciate it.

Thanks you

Start from $1$, compute its image, compute the image of this image, and so on until this goes back to $1$. Iterate starting from the smallest element not met until then.

Here $1\to5\to1$ hence your first cycle is $(15)$. Next, $2\to2$. Next, $3\to7\to9\to6\to4\to3$ hence another cycle is $(37964)$. You are left with another fixed point $8\to8$.

Thus, $\alpha=(15)(37964)$.

$1\to5\to5\to5$

$2\to1\to1\to2$

$3\to7\to7\to7$

$4\to4\to9\to3$

$5\to3\to3\to1$

$6\to6\to4\to4$

$7\to2\to2\to9$

$8\to8\to8\to8$

$9\to9\to6\to6$

The no. after the first arrow gives the image under the first cycle, 2nd arrow after the 2nd cycle and so on,

So,ultimately we have,

$1\to5$

$2\to2$

$3\to7$

$4\to3$

$5\to1$

$6\to4$

$7\to9$

$8\to8$

$9\to6$

Now you can easily figure out the cycle,

Take $1$ figure out its image ,go to the image figure out its image and so on, and when you return back to the same no. its one cycle.Go on and get all the cycles.


Limit of $\frac{x^c-c^x}{x^x-c^c}$ as $x \rightarrow c$ My question is: Show that $\lim_{x \rightarrow c} \frac{x^c-c^x}{x^x-c^c}$ exists and find its value.

Because the limit is 0/0 I've tried using L'Hopital's rule, but every time I differentiate it I still get 0/0?

We're given this hint in the question: use the fact that $a^b=\exp(b\log(a))$ and I've tried doing this to be able to differentiate and use L'Hopital's rule, and also for trying to rearrange it. Can anyone help please?

No need invoke L'Hospital here, just rewrite the fraction:
$$\frac{x^c-c^x}{x^x-c^c}=\frac{(x^c-c^c)-(c^x-c^c)}{x-c}\frac{x-c} {(x^x-c^c)}$$
By definition,

$\dfrac{(x^c-c^c)}{x-c} \;$ tends to the derivative of  $x^c$ at $x=c$, i.e. to $c^c$;
$\dfrac{(c^x-c^c)}{x-c}\; $ tends to the derivative of $c^x$ , i.e. $c^c\ln c$.
$\dfrac{(x^x-c^c)}{x-c}\; $ tends to the derivative of $x^x$, i.e.$\;c^c(\ln c+1)$.

Thus $\;\lim\limits_{x\to c}\dfrac{x^c-c^x}{x^x-c^c}=\dfrac{c^c(1-\ln c)}{c^c(1+\ln c)}=\dfrac{1-\ln c}{1+\ln c}$.

You can apply L'hopital rule, there's nothing wrong in it. Why reinvent the wheel? Lets differentiate wrt $x$:
$$\lim_{x\to c}\frac{cx^{c-1}-c^x \ln(c)}{xx^{x-1}+x^x \ln(x)} = \frac{1-\ln(c)}{1+\ln(c)}$$

Here we have used that $\frac{d(x^x)}{dx} = x\cdot x^{x-1} + x^x \ln(x)$, which can be directly evaluated (i dont remember the name of the theorem) and also by taking logarithms.


Estimating accuracy of Taylor series approximations with 2 bounds I have a question from a previous exam as such:

  Use Taylor's Inequality to estimate the accuracy of the approximation $f(x) \approx T_{3}(x)$ when $0.8 \leq x \leq 1.2$.

I computed from an earlier step in the problem that $T_{3}(x) = 1+\frac{2(x-1)}{3}-\frac{1}{9}(x-1)^2+\frac{4}{81}(x-1)^3$.

So how do I use Taylor's Inequality on this? I've only ever seen Taylor's Inequality work on one remainder term, like this.

EDIT: The original function is $f(x)=x^{\frac{2}{3}}$ centered at $a=1$.

Since $0.8=1-0.2, \; 1.2 = 1+0.2$ are symmetric around $x=1$ this is the usual situation. The error estimate from the Taylor approximation is

$$R_3 \le \frac{M |x-1|^4}{4!}$$

where $M$ is the maximum of $f^{(4)}(x)$ in the interval $(0.8,1.2),\;$ here
$M=f^{(4)}(0.8)\approx 1.455$. Therefore the error is
$$R_3 \le \frac{1.455 |0.2|^4}{4!}\approx 9.7\cdot10^{-5}$$ Plotting $f(x)-T_3(x)$ I get an actual error of $\approx 5.3\cdot10^{-5}$

Include a discussion of the accuracy of the equation by evaluating the polynomial for $t = \\{0, 1, 2, 3, 4, 5\\}$ and an example of how and why someone might use this equation.


Why is the infinite set from the axiom of infinity the natural numbers? Why is the infinite set from the axiom of infinity the natural numbers?

Is there any reason such set was chosen? Couldn't the axiom yield a set that looks like $\Bbb R$ for example?

Simplicity.

The language of set theory contains exactly one symbol, which is a binary relation symbol. That symbol is $\in$ and it is used to represent the membership relation.

In fact in sufficiently old treatments, you will find that even $=$ is dropped out of the language and is added via the conservative addition as it is definable from $\in$ when assuming the axiom of extensionality.

Now you also need to observe that the infinite set guaranteed by the axiom of infinity is not $\Bbb N$.

It is an inductive set, which may or may not be $\omega$ (the least infinite ordinal).
$\omega$ can be used to model the natural numbers internally to set theory. But we can also choose different interpretations. The important thing is that we can prove they are all "essentially the same" (read: isomorphic).

So what is $\Bbb R$? You could argue that $\Bbb R$ is the unique linear order which is both Archimedean and complete. But how would you express this in the language of set theory? You will have to express what is a linear order, what does it mean to be Archimedean (which will include invariably talking about $\Bbb N$, or $\omega$, thus postulating its existence). And what does it mean that it is unique?

And you will have to do all this work in your axioms. Axioms are supposed to be simple as possible. They should be almost definitions (and some argue that they are definitions). From the axioms of $\sf ZFC$ we can prove that we can internalize first-order logic, define semantics in a sound way, and that there exists a unique-up-to-isomorphism such and such structure or another.

Now compare this to the axiom of infinity which simply postulates the existence of an inductive set. Not even "there is a least inductive set". Just the existence of one inductive set.

This is simplicity. This is beauty.

Andres remarked that we can postulate other type of "infinite sets" existing. We can postulate that there exists a set which is not Dedekind-finite, or that there is a set which is not finite using other formulations that do not refer to the natural numbers. But usually these formulations require something like "injection" or "power set" or "maximal element". All those things require us to first interpret ordered pairs, and what does it mean for something to have certain properties. But ordered pairs can be interpreted in several different ways, not just the Kuratowski interpretation. Note that the Replacement axioms, which ultimately talk about functions, don't refer to functions as objects. They talk about formulas which define a "functional relation", without talking about the set or class of ordered pairs they can be used to define.

So again you end up with something which is more contrived, or more complex to state, and requires an extra mile of work. And again, compare this to the simplicity of the following axiom:
$$\exists A\bigg(\exists z\big(z\in A\land\forall y(y\notin z)\big)\land\forall x\big(x\in A\rightarrow(\exists z(z\in A\land\forall y(y\in z\leftrightarrow(y\in x\lor y=x))))\big)\bigg)$$
Or, if you prefer the simpler formulation after adding $\varnothing$ to the language, and using $\{\}$ notation,
$$\exists A(\varnothing\in A\land\forall x(x\in A\rightarrow x\cup\{x\}\in A)).$$

To get a set that "looks like" $\mathbb R$ (in the sense of having that cardinality), you can take $2^A = \text{the set of all subsets of }A$, where $A$ is countably infinite.  To show that that exists, you first need to know that a countably infinite set exists.


Recurrance relation with non-integer part Apologies if this has already been asked, but I'm stuck.

I have a recurrance relation that I want resolve to the form $a_n=f(n,\gamma)a_0$ but I don't know what to do.

The relation is: $$a_n=a_{n-1}\frac{n-1-\gamma}{n^2}$$ where $\gamma$ is not an integer.

There is obviously a $(n!)^2$ term on the bottom. I don't know how to deal with the top, seeing as it is not going to be an integer. I'm thinking there has to be something to do with gamma functions, but there are some tricky things that happen depending on the value of $\gamma$ you choose. (for instance, the sign flipping in the sequence if $\gamma>0$).

Please help!

For future reference of anyone who comes across this, the solution uses the Pochhammer symbol.

$$a_n=\frac{(-\gamma+n-1)_n}{(n!)^2}a_0=\frac{(-\gamma)^{(n)}}{(n!)^2}a_0$$

Where $(x)_n$ is the falling factorial and $x^{(n)}$ is the rising factorial


Orbits existing for all time For part $c)$ I understand why the above argument implies that no solution can ever tend to infinity. However I don't understand why this implies that solutions exist for all time. Why if a solution tended to infinity in the $x-y$ plane would it mean that solutions do not exist for all time?

It is true that an unbounded solution of an ODE can exist for all positive times, for instance $$\begin{cases}x'=x\\y'=0\\x(0)=y(0)=1\end{cases}$$

Has maximal solution $g(t)=\begin{pmatrix}e^t\\1\end{pmatrix}$.

What you are supposed to prove is, roughly speaking, that if a solution ceases to exist in a finite time $t_e$, then it must diverge at $t_e$.

What you need to use is the following result, which sometimes goes under the name of "Compact-Escape Theorem":

  Let $F:\mathbb{R}\times\mathbb{R}^n\rightarrow\mathbb{R}^m$ a $C^1$ function. Let $y$ a maximal solution of the ODE
  $$\begin{cases} y'=F(t,y)\\ y(0)=y_0\end{cases}$$
  Then, the orbit $\Omega_{y_0}:=\{(t,s)\in\mathbb{R}\times\mathbb{R}^m\ s.t.\ t\geq0\wedge s=y(t)\}$ is not contained in any compact subset of $\mathbb{R}\times\mathbb{R}^m$.

In your case (c), since the image of $y$ is contained in a square, its domain cannot be limited.

Take a look at the likewise quadratic ODE $y'=y^2$ which transforms by separation to
$$
\frac1{y_0}-\frac1{y(t)}=t-t_0\iff y(t)=\frac{y_0}{1-y_0(t-t_0)}
$$
These solutions have poles in finite time, thus the solution for an IVP will not exist for all time.