Is there a quadruple product rule? The triple product rule in multivariable calculus is widely used.  Can a quadruple product rule equation be written for an equation f(x,y,z,z2)=0?

Actually, there is general product rule for $n$-tuple. It is almost the direct consequence of implicit function theorem. Assume $n$ variables satisfies $F(x_1,\ldots,x_n)=0$, we have
$$\frac{\partial x_i}{\partial x_j}=-{\partial F/\partial x_j\over\partial F/\partial x_i}$$
Then multiply all fractions
$$\frac{\partial x_1}{\partial x_2}\cdots\frac{\partial x_{n-1}}{\partial x_n}\frac{\partial x_n}{\partial x_1}=(-1)^n{\partial F/\partial x_2\over\partial F/\partial x_1}\cdots{\partial F/\partial x_n\over\partial F/\partial x_{n-1}}{\partial F/\partial x_1\over\partial F/\partial x_n}=(-1)^n$$

This is what would make sense to me: $(fghi)'=f'ghi+fg'hi+fgh'i+fghi'$
You can keep adding $n$ functions to this rule.


arc length question calculus we have to calculate the arc length of the following function
$y=\sqrt{(\cos2x)} dx$ in the interval $[0 ,\pi/4]$. I know the arc length formula but following it becomes an integral thats really complex...need help....

I got the following integral $$\int_0^{\frac{\pi }{4}} \sqrt{\sin (2 x) \tan (2 x)+1} \, dx$$ no hope that this has an algebraic solution.

$\int_{0}^{\frac{\pi}{4}} \sqrt{\cos2x} dx = \frac{1}{2}\int_{0}^{\frac{\pi}{4}} \sqrt{\cos{u}} du = \frac{1}{2}\cdot \:2\text{E}\left(\frac{u}{2}|\:2\right)  = \frac{1}{2}\cdot \:2\text{E}\left(\frac{2x}{2}|\:2\right)  = \text{E}\left(x|2\right)+C$, where $\text{E}\left(x|m\right)$ is the elliptic integral of the second kind.
See: https://math.stackexchange.com/a/19786/733593


Determine whether the sequence converge or diverge The sequence is

$$a_{n+1} = 2a_n - 1$$

and

$$a_1 = 1$$

The sequence should be

$$\{1, 1, 1, 1, 1, 1, ...\}$$

and I get
$$\lim_{n\to\infty}a_n = 1$$

However
$$\lim_{n\to\infty}s_n = s = \infty$$

So is the sequence convergent or divergent? In other words, how do I judge the sequence?

$\displaystyle a_{n + 1} - 1 = 2\left(a_{n} - 1\right) \implies
a_{n + 1} = 2^{n}\left(a_{1} - 1\right) = 0\quad\mbox{because}\ a_{1} = 1$.

  In general, the sequence diverges whenever $\displaystyle a_{1} \not= 1$.

If the seq. is convergent, then $\lim_n a_{n+1}=\lim_n a_{n}$. Now for this limit, let $x=\lim a_{n+1}=\lim a_n$, then

$$
x=2x-1
$$
has solution $x=1$. Can you now prove that for every $\epsilon>0, \exists n \in \mathbb{N}$ s.t $\forall m>n$ we have $|a_m-1|<\epsilon$?


What is the name of the quadrilateral shape described by two radii and two arcs? What is the name of the four sided shape described by two radii and two concentric arcs?
Like each black and white section taken individually on this image

Or the double score / triple score areas on a dart board.

I'm attempting to create each section individually, programmatically and need help on the best thing to google to get help actually making this.

Since I'm not a native English speaker, I was a bit reluctant to post something I made up myself as an answer. But Googling a little and the comment by amWhy encouraged me (showing me that I didn't really make it up, after all):

I think the word you're looking for is

  Annular Sector

in analogy with a circular sector. I think that everyone knowing what a circular sector is, should be able to figure out what an annular sector is. An alternative word would be sector of an annulus.

You can specify an annular sector by giving its center $C$, the opening angle $\theta$ and the radii $0 \lt r \lt R$ of the two circular arcs bounding it:

Both these terms, annular sector and sector of an annulus give quite a few hits in Google. For example Googling for area annular sector brings up this link where you can find the formula

  $$\textbf{Area of an angular sector:} \qquad \qquad A_{r,R,\theta} = \frac{(R^{2} - r^2)}{2}\cdot \theta$$

for the area $A_{r,R,\theta}$ of an annular sector with radii $0 \lt r \lt R$ and angle $\theta$ (measured in radians). This is because an annular sector is the circular sector of radius $R$ minus the circular sector of radius $r$ whose areas are $\dfrac{R^2}{2} \cdot \theta$ and $\dfrac{r^2}{2}\cdot\theta$, respectively. If the opening angle $\theta$ is given in degrees, you can convert it to radians using the formula $$\operatorname{radians} = \operatorname{degrees} \times \frac{\pi}{180}.$$

The length of the circumference $c_{r,R,\theta}$ of an annular sector with radii $0 \lt r \lt R$ and angle $\theta$ is

  $$ \textbf{Circumference of an annular sector:} \qquad c_{r,R,\theta} = R \cdot (\theta + 2) + r \cdot (\theta-2)$$

where again $\theta$ is measured in radians. This is obtained by adding the lengths $R \theta$ and $r \theta$ of the two circular arcs and twice the length $(R-r)$ of the straight side.

A quadrilateral has straight sides, so I assume you are referring to the quadrilateral that connects the 4 relevant points.

It's a trapezoid. More specifically, it is an isosceles trapezoid

Since the circles from which you got the arcs are concentric, we know we get overlapping isosceles triangles when we take, separately, each of the two pairs of arc points together with the center of the circle. For the large isosceles triangle, this implies that the far side of the quadrilateral (relative to the center of the circle) makes angles with each of the radius that are each equal to the other. Similarly, the near side creates equal angles with the radii.

The overlapping isosceles triangles have the same central angle, so the 4 base angles are all (180-A)/2 degrees. This same configuration (ultimately that all corresponding angles where a radius intersects the near and far sides are equal) is defined whenever we have parallel lines intersecting a third line. Thus, we see that the near and far sides are parallel. This means we have a trapezoid. [Note that the bases are of different sizes to match the isosceles triangles which are definitely not congruent, so we don't have a rectangle.]

So a trapezoid with opposite base angles being congruent is an isosceles trapezoid.


What does it mean when a differential form "stays the same"? For example, consider the differential one-form
$$\frac{\mathrm dw}{1-w^2}$$
If we make the change of coordinates $w=1/z$ then we see that
$$\frac{\mathrm dw}{1-w^2} \longrightarrow \frac{\mathrm dz}{1-z^2}$$
Is there any significance to the form "being the same" in both coordinate systems?

Context: I'm looking at meromorphic differential forms on Riemann Surfaces.

We can think of the substitution $w\leftrightarrow w^{-1}$ as an action of the cyclic group of order $2$ on the Riemann sphere. The quotient by this action is again the Riemann sphere, and we can identify the quotient map with
$$
\begin{align}
\mathbb{CP}^1&\to\mathbb{CP}^1\\
w&\mapsto t:=w+w^{-1}.\end{align}$$
 The property that the one-form $\omega:=dw/(1-w^2)$ is invariant under $w\leftrightarrow w^{-1}$ means that $\omega$ actually comes from a one-form on the quotient. In other words, $\omega$ can be written in terms of $t$:
$$
\frac{dw}{1-w^2}=\frac{dt}{4-t^2}.
$$

Then you can define a global form by defining in each coordinate chart and patch them together.


Solutions to the diophantine equation $x^3+y^3+z^3+w^3=1$ What is known about the solutions $(x,y,z,w)\in \mathbb{Z}^4$
of the diophantine equation
$$x^3+y^3+z^3+w^3=1$$
Can you suggest me a book or a paper treating this problem?

Theorem.-The equation $x^3+y^3+z^3+w^3=n$ has an infinity of integer solutions if there exists one solution $(a,b,c,d)$ such that $$-(a+b)(c+d)\gt 0$$ is not a perfect square, and $a\ne b$, or $c\ne d$.

(See Diophantine equations, L. J. Mordell. Academic Press, p. 58).

Since $\mathbb F_7^3=\{0,\pm1\}$ necessarily one of the unknowns is a multiple of $7$ because if not we would have
$$(\pm1)+(\pm1)+(\pm1)+(\pm1)\equiv 1\pmod7$$ which is impossible.

Trying to find a particular solution I find out $(x,y,z,w)=(14,\space 30,-23,-26)$ which satisfies the conditions of the above theorem. Thus, there are infinitely many solutions which are related, according to the mentioned book, to a Pell's equation (which as it is well known has an infinity of integer solutions).

This can be solved in pieces:

$a+b+c+d=1$
$(x)\mapsto x^3$ (used 4 times)
$\mathbf{Z}^4$ integer lattice

To solve the problem, we need inverse image. The 1st step is just a bar of length 1 split to 4 parts from arbitrary places(in $\mathbf{R}$). in 2nd step, inverse of $x^3$ is just cube root $\sqrt[3]{x}$. 3rd step those cube root values need to hit the lattice points in the integer lattice. So to get solution, we need to understand how cube root behaves. There's a path $$1 \mapsto (a+b+c+d) \mapsto (a,b,c,d) \mapsto (\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c},\sqrt[3]{d}) \mapsto P$$ $$ (k_1Z_1+k_2Z_2+k_3Z_3+k_4Z_4) \mapsto P$$, where the first two $\mapsto$ generates infinite tuples satisfying (1), and the last $\mapsto$ filters them away, with $a,b,c,d \in R$ and $k_1,k_2,k_3,k_4 \in Z$. There's a pullback $R^4 \times_P Z^4$.

Note, if you want solutions out from this construct, you should remove the "diophantine" keyword, and thus the lattice disappears, and you can then get solutions from the path. Approximate solution will implement the pullback as just $(x,y,z,t)
 \mapsto (\lfloor{x}\rfloor,\lfloor{y}\rfloor,\lfloor{z}\rfloor,\lfloor{t}\rfloor)$, where there's $\pm 1$ error in the results for each element.


Given a matrix $A$ with $\operatorname{tr} (A) = 0$, prove that there is a B such that $\forall 1\leq i\leq n :(B^{-1}AB)_{i,i}=0$ I've tried using some matrices $B^{-1}$ that switch the rows, but the $B$ at the end placed the elements back in the diagonal (in different order) so I couldn't find a rule.

Lemma
Given any matrix $A$ of order at least $2$, there exists an invertible matrix $E$ such that $(E^{-1}AE)_{11} = 0$.

Proof: Let $A = [a_{ij}]_{n \times n}$, $n \ge 2$. If $a_{11} = 0$, the result holds with $E = I$, the identity matrix of order $n$. If $a_{12} = 0$, then let $E$ be the permutation matrix obtained by interchanging the first and second rows of $I$. Then $E^{-1}AE$ is the result of interchanging the first and second rows as well as first and second columns of $A$, and therefore has $(1,1)$ entry zero.

Suppose $a_{11}$ and $a_{12}$ are non-zero. Define the $n \times n$ matrix
\begin{equation*}
E = \left[\begin{array}{cc|c}
a_{12} & 0 & \mathbf 0_{1 \times (n-2)}\\
-a_{11} & 1 & \mathbf 0_{1 \times (n-2)}\\
\hline
\mathbf 0_{(n-2) \times 1} & \mathbf 0_{(n-2) \times 1} & I_{n-2}
\end{array}\right].
\end{equation*}
The inverse of $E$ is easily verified to be
\begin{equation*}
E^{-1} = \left[\begin{array}{cc|c}
1/a_{12} & 0 & \mathbf 0_{1 \times (n-2)}\\
-a_{11}/a_{12} & 1 & \mathbf 0_{1 \times (n-2)}\\
\hline
\mathbf 0_{(n-2) \times 1} & \mathbf 0_{(n-2) \times 1} & I_{n-2}
\end{array}\right].
\end{equation*}

Then the $(1,1)$ entry of $AE$ is $a_{11} a_{12} - a_{12} a_{11} = 0$. Multiplication on the left by $E^{-1}$ results in a matrix with the first row elements multiplied by $1/a_{12}$ (and other rows affected in various ways). Therefore, $E^{-1}AE$ also has $(1,1)$ entry zero. $\qquad \square$

Now, we know that similar matrices have the same trace, so $\operatorname{trace}(E^{-1}AE) = \operatorname{trace}(A)$. Thus we have the following corollary.

Corollary
Every square matrix of order at least $2$ is similar to a matrix with the same trace, and having $(1,1)$ entry zero.

Proposition
Every square matrix $A$ is similar to a matrix with the last diagonal entry equal to $\operatorname{trace}(A)$, and all preceding diagonal entries zero.

Proof: We prove by induction on $k$, $1 \le k \le n - 1$, that $A$ is similar to a matrix with the first $k$ diagonal entries $0$, and having the same trace as $A$ due to similarity. Then the result follows by letting $k = n - 1$.

From the above corollary, the result is true for $k = 1$. That is, there is a matrix similar to $A$ and having the same trace as $A$, with its first diagonal entry equal to zero. Suppose the result to be true for $k$. Thus, without loss of generality, assume that the first $k$ diagonal entries of $A$ are all zero. Let $A$ be partitioned as given below (with matrices $X$ and $Y$ of appropriate sizes).
\begin{equation*}
A = \left[\begin{array}{c|c}
Z_{k \times k} & X\\
\hline
Y & B_{(n - k) \times (n - k)}
\end{array}\right]
\end{equation*}
Then $Z$ has a zero diagonal, and $\operatorname{trace}(B) = \operatorname{trace}(A)$.

Now, by the above lemma, we have a matrix $E$ of order $n - k$ such that $E^{-1}BE$ has $(1,1)$ entry zero. Define an $n \times n$ matrix
\begin{equation*}
F = \left[\begin{array}{c|c}
I_k & \mathbf 0'\\
\hline
\mathbf 0 & E
\end{array}\right]
\end{equation*}
with the zero matrices $\mathbf 0$  and $\mathbf 0'$ of sizes same as those of $X$ and $Y$ respectively.

Then
\begin{align*}
F^{-1}AF & = \left[\begin{array}{c|c}
I_k & \mathbf 0'\\
\hline
\mathbf 0 & E^{-1}
\end{array}\right]
\left[\begin{array}{c|c}
Z& X\\
\hline
Y & B
\end{array}\right]
\left[\begin{array}{c|c}
I_k & \mathbf 0'\\
\hline
\mathbf 0 & E
\end{array}\right]
\\
& = \left[\begin{array}{c|c}
I_k & \mathbf 0'\\
\hline
\mathbf 0 & E^{-1}
\end{array}\right]
\left[\begin{array}{c|c}
Z & XE\\
\hline
Y & BE
\end{array}\right]
\\
& = \left[\begin{array}{c|c}
Z & XE\\
\hline
E^{-1}Y & E^{-1}BE
\end{array}\right]
\end{align*}

Thus, $F^{-1}AF$ is a matrix similar to $A$ (and hence with zero trace) with the first $k + 1$ diagonal entries zero (since the first diagonal entry of $E^{-1}BE$ is zero). By induction, the result is true for all $k \le n - 1$. $\qquad \square$

As a special case of the above proposition, if $\operatorname{trace}(A) = 0$, then $A$ is similar to a matrix with the last diagonal entry as well as all preceding diagonal entries zero. Thus we have the main result.

Theorem
Every matrix $A$ with zero trace is similar to a matrix with zero diagonal.

Hint:
For a $2 \times 2$ matrix, consider one with $1$ and $-1$ on the diagonal. (Note that there cannot be any off-diagonal elements here, because the evalues differ). Then picking $B$ to be rotation by $\pi/4$ i.e., the matrix
$$
s\begin{bmatrix}
1 & -1 \\
1 & 1
\end{bmatrix},
$$
where $s = \frac{\sqrt{2}}{2}$,
produces a matrix of the desired form, and this clearly generalizes to other eigenvalues besides $1$.

How'd I find this? I looked for a vector $v$ that the original matrix sent to a vector $w$ with $v \cdot w = 0$, since that's what helps produce a 0 on the diagonal. Since the original matrix simply reflects through the $x$-axis, the vector $(1, 1)$ goes to $(1, -1)$, which is perpendicular to it. When I found two such independent vectors (a similar argument applies to $(1, -1)$, for example), I chose them as my columns for $B$ (i.e., I represented the original transform in that basis).  (Along the way, I made them unit vectors so that it'd be easier to compute $B^{-1}$.)

For $3 \times 3$, there are three cases to consider: diagonal with three distinct e-values that sum to zero, diagonal with entries $a, a, -2a$, and the same thing, but with a $1$ in the superdiagonal entry above $a$ and $a$.

In each case, you need to do a similar analysis/discovery of vectors.


lim calculus problem with infinity $$\lim_{n→∞} \left( \left(\frac{n+3}{n+5}\right)^{n+4} + \sqrt[2n]{3n}\right)$$

I tried replacing n+5 with u and simplifying but the answer seems off. Help would be greatly appreciated.

We need to do some rewriting:

$$\begin{align}
\left(\frac{n+3}{n+5}\right)^{n+4} &= \left(\frac{n+5-2}{n+5}\right)^{n+4} \\
&= \left(1-\frac{2}{n+5}\right)^{n+5-1} \\
&= \underbrace{ \left(1-\frac{2}{n+5}\right)^{n+5} }_{\to e^{-2}}\ \underbrace{ \left(1-\frac{2} {n+5}\right)^{-1} }_{\to 1}
\end{align}
$$

So the left term goes to $e^{-2}$. The right side can be written as $(\sqrt[n]{3n})^{\frac12}$. The $n$-th root of any polynomial in $n$ goes to $1$, so this goes to $1$. The overall limit is $e^{-2}-1$.

The inner fraction simplifies to 1, so 1 to infinity is 1. 1 + infinity = infinity.

(then again it is quite difficult to understand from the formatting)

P.S. After replacing u = n+5, you get $\frac{u - 2}{u}$, which isn't off.


Does n power of e grow much more faster than its Maclaurin polynomial? I wonder how to calculate the following limit:
$$
\lim_{n\rightarrow\infty}\frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}}
$$
In the first sight, I think it should be zero, because exponential function is much faster than polynomial. But the upper of the expression is the Maclaurin polynomial of $e^{n}$. With the growth of n, it approaches to $e^{n}$. Consider there is a roughly way to estimate the remainder of $e^{n}$ rather than
$$
R_{n+1}(n)=\frac{\xi^{n+1}}{(n+1)!}     \ for  \ \ \xi\in(n,+\infty)
$$
Because
$$
\frac{1+n+\frac{{}n^{2}}{2!}+\cdots +\frac{n^{n}}{n!}}{e^{n}}=1-\frac{R_{n+1}(n)}{e^{n}}
$$
But it's hard to continue.

Let a be the difference error between $e^n$ and maclaurins series the value of the expression now becomes $1-\frac a{e^n}$ and now as n tends to $\infty $ the difference error a tends to zero and $e^n$ tends to $\infty$ and therefore the limit tends to 1 which implies that they both grow at same rate.


Determine when the arc-length is (in)finite. Problem:
Let $\phi : [0,1] \to \mathbb C$ be given by
$$\phi(t) := t + it\sin\left(\frac{\pi}{t^p}\right)$$
for $t \in (0,1]$, and define $\phi(0) := 0$. Then $\phi$ is continuous. For what values $p > 0$ is the arc-length finite? Infinite?

Attempt: Recall that arc-length is given by
$$AL(\phi) := \int_0^1 |\phi'(t)|\,\mathrm{d}t$$
It turns out that a direct computation of arc-length is absolutely miserable. Instead, I tried various ways of bounding it. For example, using that
$$\sqrt{1 + x^2} \leq 1 + |x|$$
This yields, at best, an opportunity to Taylor expand, but doing this led me to believe that this particular approximation is not even convergent. Unfortunately, I'm absolutely stumped!

To determine the values of $p$ for which the arc length is integrate, I was hoping to use the inequality
$$\mathrm{AL}(\phi) \geq \frac{\left|\int_{\mathrm{Im}(\phi)} f(z) \,\mathrm{d}z\right|}{\max\limits_{z \in \mathrm{Im}(\phi)} |f(z)|}$$ with a clever choice of $f$. But I just haven't been clever enough yet!

Let $\theta = \pi t^{-p}$. After some work, we can show that the arclength is given by $$K_p\int_\pi^\infty \sqrt{1 + (\sin \theta - p\theta \cos \theta)^2} \frac{d\theta}{\theta^{1+1/p}}$$
where $K_p$ is some constant depending on $p$ for which all we care about is that it is finite. Dropping the unimportant constant, we can further change it to $$\int_\pi^\infty \sqrt{\frac{1 + \sin^2 \theta}{\theta^2} - \frac{p\sin 2\theta}{\theta} + p^2\cos \theta^2} \frac{d\theta}{\theta^{1/p}}$$

For sufficiently large $\theta$, the first two terms are less than $p^2$, so we have $$\sqrt{\frac{1 + \sin^2 \theta}{\theta^2} - \frac{p\sin 2\theta}{\theta} + p^2\cos \theta^2} \le \sqrt{p^2+ p^2\cos \theta^2} = p|\sin \theta| \le p$$ which is sufficient to handle the case of $p < 1$, and should hopefully suggest ways of looking at $p \ge 1$ as well.

See (fix the url): http://i.stack.imgur.(com)/w6bEr.png

Given any of the oscillations of phi you can get an upper and lower bound on the arclength of the oscillation by considering these red and blue curves in this quickly-made image. Picking the most natural points for where the red and blue curves peak gives nice bounds so that once you sum over all oscillations, you manage all p-cases at once.


Proof of Alternate, Corresponding and Co-interior Angles During school our teacher always explains the proof for all theorems even simple ones such as why does the angles in a triangle of add up to $180$ and they all involve alternate, corresponding or co-interior angles. However it has never occurred to me that he never shown the proof of how are alternate angles are equal.

Could any explain the proof for alternate, corresponding and co-interior angles?

Not everything has a proof. As sawarnik link says that there are axioms and postulates regarding this fact. We havent been able to find 2 parallel lines with transversal who's alternate and co interior angles arent equal :)!
The day someone does euclid and others will be proved wrong. But anyways u can always geometrically prove this by making a parallel line transversal setup. Cut the co interior angles briefly and putg them next to each other you will see that they will be exactly like a protractor or semi circle making 180° similarly corresponding  interior and alternative interior angles will overlap proving that they are equal...If they aren't you will become famous!!

The Corresponding Angles Postulate is the base for proving the other cases.

http://hotmath.com/hotmath_help/topics/corresponding-angles-postulate.html

Also go to: http://www.cliffsnotes.com/math/geometry/fundamental-ideas/postulates-and-theorems

As a postulate it does not have a proof and is part of the Parallel Lines axiom.

To summarize: based on the corresponding angles postulate you may prove the theorems with regard to alternate and co-interior angles. From there you derive the sum of the triangle angels to be 180.

Suggest you take advantage of the wealth of knowledge on WWW


Integration of $\sqrt{ x^2 - x\cdot \sin(x) + 4}$ How to integrate

$$\sqrt{ x^2 - x\cdot \sin(x) + 4}$$

with respect to $dx$?

I tried solving this using substitution,  but it doesn't seem to work.  Can somebody give me a little direction on how to  proceed.

You can try this substitution below. From the point I stopped on, it should be a matter of organizing things and doing basic stuff.

$$\eqalign{
  & \int {\sqrt {{x^2} - x\sin x + 4} } dx = \int {\sqrt {4\left( {{{{x^2}} \over 4} - {{x\sin x} \over 4} + 1} \right)} } dx = 2\int {\left( {{{{x^2}} \over 4} - {{x\sin x} \over 4} + 1} \right)} dx  \cr
  & x = 2\tan \theta   \cr
  & {{dx} \over {d\theta }} = 2{\sec ^2}\theta   \cr
  & 2\int {\left( {{{4{{\tan }^2}\theta } \over 4} - {{4ta{n^2}\theta } \over {4\sqrt {4 + 4{{\tan }^2}\theta } }} + 1} \right)} 2{\sec ^2}\theta d\theta   \cr
  & 2\int {\left( {{{\tan }^2}\theta  - {{ta{n^2}\theta } \over {2\sec \theta }} + 1} \right)} 2{\sec ^2}\theta d\theta  \cr} $$

(sorry if I left anything wrong in the middle of all that!)


Why isnt $\lim_{x\to 0}\frac{\sin x}{x} = 0$? $$\lim_{x\to 0}\frac{\sin x}{x}$$

Can't we say this:

$$\lim_{x\to 0}\left[\frac{1}{x}\right]\cdot\lim_{x\to 0}[\sin x]$$

So it would be infinity times 0, which would be 0.

Why is this method wrong?

Because the rule that you are using, that:

$$\lim a_n b_n = \lim a_n \lim b_n$$

only works if the limits exist. For example here is a screenshot straight from the wikipedia page:

Notice how it says, "provided the limits on the right side of the equation exist". In your example,
$$\lim_{x \to 0}{1 \over x}$$
does not exist so you can't simply apply the limit laws. Something more refined like l'Hôpital's rule or a more detailed analysis is required (depending on how you defined the trig functions/proved l'Hôptial's).

If you would like a bit of a visual intuition as to why $$\lim_{x \to 0} {\sin x \over x}= 1$$

here is a jank gif I made to show it. I believe one of the other answers discusses this analytically but notice how when you zoom in enough at 0, the graphs of $\sin x$ and $x$ effectively become the same? This is why the ratio of their limits as they approach $0$ goes to $1$.

Maybe as an exercise, compare this to how $\cos(x) - 1$ looks near $x = 0$ and then try and hypothesize an answer to $$\lim_{x \to 0} {\cos x - 1 \over x}$$

This limit has $\frac{0}{0}$, so L'Hopital's rule is admissible.  Taking the derivative of both the top and bottom, we get $$\lim_{x \to 0} \ \cos(x) = 1$$
In essence, when we see a case of $\frac{0}{0}$ or $\frac{\infty}{\infty} $, it is equivalent to take the derivative of both the numerator and denominator to guage the relative speed at which the two respectively approach those values. If you would like to see an argument that does not involve calculus, you can easily look up a geometric proof using the squeeze theorem on Google.


Help with Inverse Function and Composition of Functions? I'm currently doing work on discrete mathematics in my free time and am having some difficulties with understanding some questions pertaining to Relations and Functions. To be specific, I'm stuck on the following practice question:

Let P1 = (3 4 1 2 5), P2 = (3 5 1 2 4) and P3 = (5 1 4 2 3) be three
permutations

1.) Find the permutations below:
    a.) P3^-1
    b.) (P3 ◦ P1)^-1
    c.) P1^-1 ◦ P2

2.) Show that P1 ◦ P2 != P2 ◦ P1

Starting with the easy stuff I know P3^-1 would be (2 4 5 3 1). But after that point I'm not sure what to do to determine the other answers. Any help is appreciated! Thank you.

To find the inverse of a permutation that is a cycle all we have to do is write the elements of the cycle in reverse order. Thus the inverse of (1 2 3 4) is (4 3 2 1). Since a cycle can be written with any of its elements as the first term we can also write this inverse as (1 4 3 2). This gives an alternative way to write down the inverse of a cycle. Fix the first element in the cycle and write the remaining elements in reverse order. Thus, the inverse of (1 2 3 4 5) is (1 5 4 3 2).

so $P_3^{-1}=(3,2,4,1,5)=(1,5,3,2,4)$

$ (P_3 ◦ P_1)=(5 1 4 2 3)◦(3 4 1 2 5)=(1,3,2)\Rightarrow (P_3 ◦ P_1)^{-1}=(1,2,3)=(2,3,1) $....


Definition of linking number for disjoint submanifolds of the sphere- A problem from Milnor's book In problem number 13 of Milnor's 'Topology from the Differentiable Viewpoint', the linking number for two compact boundary-less manifolds $M,N \subset \mathbb{R}^{k+1}$ of dimensions $m,n$ such that $m+n =k$ is defined as the degree of the map $\lambda\colon M \times N \to S^{k}$ given by $\lambda(x,y)=\frac{x-y}{\|x-y\|}$.
The problem then asks to prove some properties of the linking number. The final question, where I need help, is

  Define the linking number for disjoint manifolds of the sphere $S^{m+n+1}$.

I have no clue as to what the definition should be. I need this in order to proceed to the subsequent problems. Could somebody please give the definition, and a suitable reference ? Thanks !

Note that $M$ and $N$ do not cover all of $S^{n+m+1}$. If the point $P \in S^{n+m+1}$ is missed, then $M$ and $N$ live in $S^{n+m+1} \setminus \{P\}$, which is homeomorphic to $\mathbb{R}^{n+m+1}$. You can now use your old definition.

Note that, although this definition seems to depend on the choice of $P$, it does continuously so. Since the linking number is a (discrete) integer, it means it is independent of $P$.

you can find three different definitions
in bott and tu's book: differential forms in algebra topology page 229. The definitions by intersection number and differential forms are both in sphere.


A maximal Hoeffding's inequality? Let $X_1, \cdots, X_n$ be real-valued independent random variables satisfying $|X_k|\le 1$ and $\mathbb EX_k=0$. Hoeffding's inequality tells us that for any $k=1,\cdots, n$ and $t>0$,
$$\mathbb P\Big( \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le 2 e^{-t^2/2}.$$

My question is whether there exists a similar bound for the maximum over $k$. More precisely:

  Question: Do there exist absolute constants $C>0$ and $A>0$ so that
  $$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{X_1+\cdots+ X_k}{\sqrt{k}} \Big | \ge t \Big) \le C e^{-t^2/A}$$
  holds for all $t>0$? If not what can we say about the left hand side?

Hoeffding's inequalities for absolute values are derived by determining first the bound for the value, and then double it to arrive at a bound for the absolute value. But the question here asks for a bound related to the maximum of the absolute value, not to the absolute value of the maximum, so a direct examination of the absolute value is needed.
To compact notation, we define $S_k=\Big | X_1+\cdots+ X_k \Big |$. Then we are examining

$$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big | \ge t\Big) = \mathbb P\Big( \bigcup_k \Big\{ \Big|\frac{S_k}{\sqrt{k}} \Big | \ge t \Big\} \Big)$$

Denote $I_Z$ the indicator function of the event $Z= \Big( \bigcup_k \Big\{ \Big|\frac{S_k}{\sqrt{k}} \Big |\ge t \Big\} \Big)$ and $I_1,..., I_n$ the indicator functions of the events, $\Big\{ \Big|\frac{S_1}{\sqrt{1}} \Big |\ge t \Big\},...\Big\{ \Big|\frac{S_n}{\sqrt{n}} \Big |\ge t \Big\} $, respectively.
Then by the properties of indicator functions
$$I_Z=\max\Big \{I_1,...,I_n\Big \}$$
Now if $I_Z =0$ it means that all $I_i,\; i=1,...,n$ are zero. If $I_Z=1$, it means that at least one of these individual indicator functions is unity, denote it $I_m$, and we have
$$I_m =1 \Rightarrow \Big|\frac{S_m}{\sqrt{m}} \Big |\ge t $$
Let $v = \text{argmax}_k \Big | \frac{S_k}{\sqrt{k}} \Big |$ and $\Big | \frac{S_v}{\sqrt{v}} \Big |$ the corresponding variable. Then in the case where $I_Z =1$ we have

$$\Big | \frac{S_v}{\sqrt{v}} \Big |\ge \Big|\frac{S_m}{\sqrt{m}} \Big |\ge t$$

Then in all cases, either when $I_Z=0$ or when $I_Z=1$ we have that, for some $h>0$,

$$I_Z \le \exp \left \{h\left (\Big | \frac{S_v}{\sqrt{v}} \Big |-t\right) \right \}$$

Therefore,

$$\mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big | \ge t\Big) = \mathbb P\Big( \bigcup_k \Big\{ \Big|\frac{S_k}{\sqrt{k}} \Big | \ge t \Big\} \Big) = EI_Z \le E\exp \left \{h\left (\Big | \frac{S_v}{\sqrt{v}} \Big |-t\right) \right \}$$

$$\Rightarrow \mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big | \ge t\Big)   \le e^{-ht} E\exp \left \{h \Big | \frac{S_v}{\sqrt{v}} \Big | \right \} \qquad [1] $$

By Hoeffding's lemma, for a random variable $Y$, with $EY=0,\;a\le Y \le b$ we have, for any real $\lambda$
$$ E\left (e^{\lambda Y} \right) \le \exp \left(\frac{\lambda ^2 (b-a)^2}{8} \right) \qquad [2]$$
In our case, we have
$$\Big | \frac{S_v}{\sqrt{v}} \Big | = \frac{1}{\sqrt{v}}\Big |X_1 +...+ X_v\Big | \le \frac{1}{\sqrt{v}}\Big (\Big |X_1 \Big | +...+ \Big | X_v\Big | \Big) \le \frac{1}{\sqrt{v}}v = \sqrt{v}$$

the last inequality by the initial assumptions. Since $v\le n$ we have
$$0\le \Big | \frac{S_v}{\sqrt{v}} \Big | \le \sqrt{n} \Rightarrow -E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) \le \Big | \frac{S_v}{\sqrt{v}} \Big | - E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) \le \sqrt{n} -E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) $$.
We now have a variable with zero expected value and bounded.  The length of the interval is  $b-a = \sqrt{n} -E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) + E\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big) = \sqrt{n} $ and $\lambda = h $.
Inserting these values into Hoeffding's lemma and simplifying we get

$$ E\exp \Big \{h\Big | \frac{S_v}{\sqrt{v}} \Big | - hE\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big)\Big \} \le \exp \left(\frac{nh^2}{8} \right) $$
$$\Rightarrow E\exp \Big \{h\Big | \frac{S_v}{\sqrt{v}} \Big | \Big \} \le    \exp \left(\frac{nh^2}{8} \right)\exp\Big \{hE\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big)\Big \} \qquad [3]$$
Note that the exponential that moved to the right hand side does not contain any random quantities that's why the expected value has disappeared. From previously we have
$$ \Big | \frac{S_v}{\sqrt{v}} \Big | \le \sqrt{n} \Rightarrow E\Big | \frac{S_v}{\sqrt{v}} \Big | \le \sqrt{n} \Rightarrow \exp\Big \{hE\Big(\Big | \frac{S_v}{\sqrt{v}} \Big |\Big)\Big \} \le \exp\Big \{h\sqrt{n}\Big \} \le \exp\Big \{h^2n\Big \} \; [4] $$

Inserting the RHS of $[4]$ into $[3]$ and back in  $[1]$ we obtain

$$ \mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big |\Big )= \mathbb P\Big( \Big | \frac{S_v}{\sqrt{v}} \Big | \ge t\Big) \le \exp \left(-ht +\frac{9nh^2}{8} \right) \qquad [5]$$
Minimizing the RHS over $h$ we obtain $h^* = \frac {4}{9n}t$ and inserting into $[5]$ we finally obtain

$$ \mathbb P\Big( \max_{1\le k\le n} \Big | \frac{S_k}{\sqrt{k}} \Big |\Big )=\mathbb P\Big( \Big | \frac{S_v}{\sqrt{v}} \Big | \ge t\Big) \le \exp\Big \{-\frac{2}{9n}t^2\Big \}  \qquad [6] $$

...which is a bound related to the number of r.v.'s involved.

Look at Lemma 1 in this paper:
https://arxiv.org/pdf/1312.7308.pdf
Since you are taking a maximum the bound is actually a bit different, and the exponent is not with $t^2$ but $t^2/\log(t)$.
The idea for how to prove it is first take a naive union bound over an a geometric series $k=1,2,4,8,16$ (not necessarily with 2 as the base), then use a maximal inequality that doesn't involve the denominator of $\sqrt{k}$ (see e.g Thm 20.20 here: http://www.math.wisc.edu/~roch/grad-prob/gradprob-notes20.pdf) to bound the remaining $k$ values


About the proof of a graph is not Hamiltonian. Given the following graph:

https://i.stack.imgur.com/fg2Q9.png

Is this graph Hamiltonian or not?

The answer is no. What I tried to prove is by using the fact that: "if a vertex in the graph has degree two, then both edges that are incident with this vertex must be part of any Hamilton circuit." Following that we know, edges 11,6,4,12 must be in any circuit and the edges 17,18,19 and 9,15 must be in any circuit. Knowing that, what should I do to conclude that the graph is not Hamiltonian?

Edges $18$ and $19$ are incident on vertex $m$ and must be in any Hamiltonian circuit.  But then none of the other edges from vertex $m$ are in the circuit, therefore the
circuit can't contain any edges connecting vertices $m, g, n$ to the rest of the graph.

Let G be any graph. If G is Hamiltonian, then κ(G) ≥ 2.
Since κ(G) = 1, with κ-set = {m},  then G must NOT be Hamiltonian.


Show that the area of a square is greater than the area of a rectangle. It is a common problem that : Show that the square is a rectangle of maximum area for a given perimeter and a rectangle of minimum perimeter for a given area.
It is an easy problem when we use calculus. It can done by using maxima & minima but how we shall show it using inequality. I can do it using calculus by how I show it using basic inequality. Somebody help me .

One problem is just the dual of the other one, so it is enough to prove one of them to get them both for (almost) free. Assume that we have a fixed perimeter $(2p)$ and we want to maximize the area. Our problem can be written as
$$ \max_{\substack{a+b=p\\ a,b\geq 0}} ab = \large{?} $$
and we may exploit the AM-GM inequality, stating that if $a,b\geq 0$,
$$ \sqrt{ab}\leq\frac{a+b}{2} $$
with a strict inequality if $a\neq b$. It follows that the maximum area ($p^2/4$) is achieved by $a=b$, i.e. by a square.

Assuming the same perimeter: Cauchy–Schwarz inequality


$(0,1), [0,1), [0,1]$ are not homeomorphic I need to show $(0,1)$, $[0,1)$ and $[0,1]$ are not homeomorphic using  intermediate value theorem (without using connectedness).
I have already did proved that $(0,1)$, $[0,1]$ are not homeomorphic but I struggle with the 2 other couples.
My proof: assume there is an homeomorphism $f:(0,1)\rightarrow[0,1]$ and take $a,b$ such that $f(a)=0, f(b)=1$. So using intermediate value theorem we can say that $f([a,b]) = [0,1]$. so $f$ isn't injective.
Will appreciate any help

Suppose $f:(0,1)\to [0,1)$ is a homeomorphism.
Then $\exists x\in(0,1)$ such that $f(x)=0$. By continuity for every  $\epsilon >0$, $ \exists y<x<z$ such that $\{f(y),f(z)\}\subset[0,\epsilon)$.
Without loss of generality assume that $f(y)<f(z)$.
Now by intermediate value property of $f$, we have $f$ on $(x,z)$ attains all the values in $(f(x),f(z))$. So there exists $z_0\in (x,z)$ such that $f(z_0)=f(y)$.
Contradicts the fact that $f$ is one-one.

Observe that if $\;f:X\to Y\;$ is a homeomorphism between two toplogical spaces and $\;x\in X\;$ then
$$\;g:X\setminus\{x\}\to Y\setminus\{f(x)\}\;,\;\;g(a)=f(a)\;\;\forall\;a\in X\setminus\{x\}\;$$
is also a homeomorphism on the respective top. spaces with the induced topology, and thus: if
$$f:[0,1)\to(0,1)\;\;\text{is a homeomorphism, then}\;\;g:(0,1)\to[(0,1)\setminus\{f(0)\}$$
is also a homeomorhpism. but this can't be since $\;(0,1)\;$ is (path) connected whereas $\;[(0,1)\setminus\{f(0)\}\;$ isn't.
Complete the other pair now (there you have path connectedness if you don't want connectedness...I don't understand why, but whatever)


Is the set of Natural Numbers equal to first infinte ordinal? (Note: Most of what I know about transfinite ordinals comes from this Wikipedia page.)
According to Wikipedia, each ordinal number is defined by the set of those smaller than it. That is, $0=\{\}$, $1=\{0\}$, $2=\{0,1\}$, $3=\{0,1,2\}$. However, doesn't this mean that ℕ (the set of natural numbers) is an ordinal, since it is equal to $\{0,1,2,3,4,5,6,...\}$? Since Wikipedia says the smallest infinite ordinal is $ω$, does that mean $\Bbb N = ω$ ? If it is, then why is the ω symbol used at all if it's just $\Bbb N$? If not, then what did I do wrong?
Wikipedia also says that ω is the "order type" of $\Bbb N$. I don't know what that means; is it related to my question?

Yes and no.

There's one sense in which the answer is unambiguously yes: Within the most popular scheme for embedding all of ordinary mathematics inside axiomatic set theory, the set that represents the set of natural numbers is identical to the set that represents the first infinite ordinal -- and by the way also identical to the set that represents the cardinal number $\aleph_0$.

Set theorists usually work in a context where this formalization is taken for granted, so they tend to say that $\mathbb N$ is the ordinal $\omega$. That works well for them. (They rarely use the symbol $\mathbb N$ anyway, preferring $\omega$ even in contexts where other mathematicians would use $\mathbb N$).

The arguments for an answer of no are a bit softer and more philosophical. But a good argument can be made that in everyday (rigorous but not formalized) mathematics, neither the natural numbers nor the ordinal numbers "are really" sets.

The number $2$, for example, is a separate idea from a set: namely the property shared by all collections that "contain one more than one thing". Ordinary mathematics allows these ideas -- which are not themselves sets -- to be collected as a set, which we call $\mathbb N$. And the ordinal number $\omega$ is not a set, but instead the abstract idea of the "thing" that is common about all infinite well-ordered sets that have no infinite proper prefixes.

With this "everyday" view $\mathbb N$ and $\omega$ are not the same thing, and are not even the same kind of thing.

This view is by no means shared by all "everyday mathematicians", though. Many seem to accept at least in principle to consider the set-theoretic formalization to be "what is really going on" -- but they still usually write things as if $\mathbb N$ and $\omega$ are different things, considering this distinction a helpful hint to the reader about which way to think about the thing in question is most relevant in each case.

Note that it is somewhat rare in "everyday mathematics" -- that is, outside axiomatic set theory -- to speak about ordinals at all.  So a quick first-approximation rule of thumb could be to write $\omega$ whenever you writing for set theorists and $\mathbb N$ whenever you're writing everyday mathematics; then you will usually not go wrong.

The finite ordinal numbers, the finite cardinal natural numbers are all use the same symbols. Natural numbers are used for counting finite things, cardinal numbers are used for measuring how many objects something can hold, and ordinals numbers are used for specifying which part of a list something is located at. The reason you need two different number systems when you expand to the infinite is because you can extend a list past the natural numbers and the list will still contain the same amount of objects as the natural numbers.


Showing that $\sqrt{1+x}=1+\tfrac12x+o(x)$ I have found that a nice way to explain to students what the derivative is the number $A$ such that
$$f(x+dx)=f(x) + A\,dx+o(dx),$$
motivated by the idea of approximating $f$ around $x$.
If we take this as the starting point of what differentiation is, it's easy to obtain things like $(2+dx)^2 \approx 4 + 4\,dx$ for small $x$, and so on for more general polynomials. But it would be nice to give a more non-trivial example. It feels like the next natural step is something like
$$\sqrt{1+dx} = 1 + \tfrac12dx+o(dx),$$
but how can I show this without using "heavy machinery"? At this point I've only started the topic, and my students have only seen $\epsilon$-$\delta$ style proofs of some basic limits (plus I mentioned some of the standard ones like $\sin x/x\to 1$ as $x\to 0$).
This is essentially an easy case of the more general power rule for derivatives: $d(x^r)=rx^{r-1}\,dx$, but all proofs I've seen use logarithms or the inverse function theorem (changing $f(x)=x^{1/2}$ to $f^{-1}(y)=y^2$ etc.). Is there a more direct way I can show that
$$\lim_{dx\to0}\frac{\sqrt{1+dx}-(1+\tfrac12\,dx)}{dx}=0?$$
Or maybe $\sqrt{1+x}$ is not the most natural "next step" up from polynomials, and there is another easier example I can give?


Find all positive integers $n$ for which $1372n^4 - 3 $ is an odd perfect square. Find all positive integers $n$ for which $1372\,n^4 - 3$ is an odd perfect square.

I tried $\bmod ,4,5,7$ and failed. Next, I used Vieta’s Theorem and failed again.
Any hints, please. Thank you very much!
Edit number and parity already. Sorry for typo
Edit 2 : This question is related to this question.

The equation $y^2=1372x^4-3$ has only one positive integral solution for $x$ and $y$ at which is found at $(1,37)$.
We can use the general technique in this answer https://mathoverflow.net/a/338108 to convert your quartic into Weierstrass form and then we can use MAGMA to find all integral points on the curve.
Step 1: Quartic to Cubic (Weierstrass form)
$y^2=1372x^4-3$ can be transformed into $Y^2=X^3-4116X$ using $X:=1372x^2$ and $Y:=1372xy$ via the steps below
Take
$$y^2=1372x^4-3$$
Multiply both sides by $1372^2x^2$
$$1372^2x^2y^2=1372^3x^6-3\times1372^2x^2$$
$$(1372xy)^2=(1372x^2)^3-(3\times1372)(1372x^2)$$
$$Y^2=X^3-4116X$$
Step 2: Search for Integral Points
Then using MAGMA (An online version is here for you to confirm my work for yourself: http://magma.maths.usyd.edu.au/calc/) we can run the following two lines of code to find all of the integral points on our curve:
E := EllipticCurve([0,0,0,4116,0]);
IntegralPoints(E);

And we get the result: $(0 : 0 : 1)$ which tells us that the only one solution exists (the one that we found manually $(1,37)$).
Alternatively: Easier Solution
We could also run the following to get this answer directly (I realized this command existed after doing the work above, but it confirms the same answer).
IntegralQuarticPoints([1372, 0, 0, 0, -3]);

which gives the only positive output as $[ 1, 37 ]$


Finding x given that $\lim_{n\to\infty } x^{x^{n-1}}=5$ Given a sequence $\{a_n\}^\infty_{n=1}$ defined in the following way:
$ a_1=x, \ a_2=x^x, \  a_3=x^{x^{x}}, a_4=x^{x^{x^{x}}} ... \ (x>0)$
what would be the value of $x$ if  $\lim_{n\to\infty } a_n=5$

It looks to me that you mean tetration. It is commonly written as $^nx$

You mean what is the limit for $lim_{n \to \infty}\,^nx$

That limit only exists for $0 \leq x \leq e^{e^{-1}}$

If $0 \leq x < e^{-e}$  then the limit is a periodic square wave function, whose max and min values are the solution to $x^{x^c}=c$ (x is the constant, and c the limit). You have 2 limit values, $c_1$ and $c_2$.

If $ e^{-e} \leq x \leq e^{e^{-1}}$  then the limit is the solution to the equation $x^c=c$. The limit c is calculated using the lambert W function.

${}^{\infty}x = x^{x^{\cdot^{\cdot^{\cdot}}}} = c = \frac{\mathrm{W}(-\ln{x})}{-\ln{x}} $

Given that $x=e^{e^{-1}}$ is the maximum value for which you have a limit, there is no limit for $^nx=5$.

But there is a limit for $lim_{n \to \infty} \,^{-n}x=5$, if $^0x$ is defined as $^0x>lim_{n \to \infty} \,^{n}x$. In such case, $x\approx 1,3797296901083$

That's because there is some arbitrariness in defining $^0x$. It is commonly defined as $^0x=1$, but that's just a (sensible) choice.

Yet $^nx$ is not $x^{x^{n-1}}$ (as you wrote in the title). The solution to $\lim_{n\to\infty } x^{x^{n-1}}=z$ is z because for the solution to exist, it should be $z^z=z$. But z=5 does not meet the requirements.

Let

$$\zeta = x^{x^{x^{...}}}$$

We know that

$$\zeta = 5$$

Since:

$$\zeta = x^{x^{x^{...}}}$$

We can say that:

$$\zeta = x^{\zeta}$$

Thus, you can solve for x using $5 = x^5$.


How to solve $\frac{\mathrm{d}^4 y}{\mathrm{d} x^4}-2{\frac{\mathrm{d^3}y}{\mathrm{d} x^3}}+2{\frac{\mathrm{d^2}y }{\mathrm{d} x^2}}+y=0$? I tried to solve the equation

$$\frac{\mathrm{d}^4 y}{\mathrm{d} x^4}-2{\frac{\mathrm{d^3}y}{\mathrm{d} x^3}}+2{\frac{\mathrm{d^2}y }{\mathrm{d} x}}+y=0$$

by writing its Auxiliary Equation

$$
m^{4}-2m^{3}+2m^{2}+1=0,
$$

But now I am unable to solve this polynomial.

go to wolfram alpha
https://www.wolframalpha.com/input/?i=complex+z+%3A++z%5E4%E2%88%922*z%5E3%2B2*z%5E2%2B1%3D0
if you don't want to use the awful formula


Does the improper integral $\int_0^\infty e^{-x^2}dx$ converge? I want to show the convergence of the following improper integral $\int_0^\infty e^{-x^2}dx$.
I try to use comparison test for integrals
$x≥0$, $-x ≥0$, $-x^2≥0$ then $e^{-x^2}≤1$. So am ending with the fact that $\int_0^\infty e^{-x^2}dx$ converges if $\int_0^\infty dx$ converges but I don’t  appreciate this. Thanks

Write

$$\int_0^\infty e^{-x^2} \, dx = \int_0^1 e^{-x^2} \, dx + \int_1^\infty e^{-x^2} \, dx$$

The first integral on the right hand side is of a bounded function over a bounded interval, hence finite. For the second notice that we have $x^2 > x$ when $x > 1$. Therefore, $e^{-x^2} < e^{-x}$ for $x > 1$ and it follows that

$$\int_1^\infty e^{-x^2} \, dx < \int_1^\infty e^{-x} \, dx$$
$$= \lim_{x \to \infty} -e^{-x} + e^{-1} = 1/e < \infty$$

A simple answer as @neemy… Remember the function of Normal probability

$$N(\mu,\sigma)=\displaystyle{\int_{-\infty}^{+\infty}{\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{1}{2}\left(\frac{x-\mu}{\sigma}\right)^2}\;dx}=\bf{1}}$$

For $\mu=0$ and $\sigma=1$
$$\displaystyle{\int_{-\infty}^{+\infty}{\frac{1}{\sqrt{2\pi}}e^{-\frac{x^2}{2}}\;dx}=\bf{1}}$$

Introduce a change of variable with its respective Jacobian
$$x=\sqrt{2}\;u\\dx=\sqrt{2}\;du$$

and result
$$\displaystyle{\int_{-\infty}^{+\infty}{\frac{\sqrt{2}}{\sqrt{2\pi}}e^{-u^2}\;du}=\bf{1}} \Rightarrow \displaystyle{\int_{-\infty}^{+\infty}{e^{-u^2}\;du}=\bf{\sqrt{\pi}}}$$

The Normal is even function, therefore
$$\displaystyle{\int_0^{+\infty}{e^{-u^2}\;du}=\frac{\sqrt{\pi}}{2}}$$

P.D.: Excuse my English, please.


A square sheet of paper ABCD is so folded that B falls on the midpoint M of CD.In which rtio the crease will divide BC. A square sheet of paper $ABCD$ is so folded that $B$ falls on the midpoint $M$ of $CD$. In which ratio will the crease divide $BC$?

Let the crease be $QR$, with $Q$ on $BC$ and $R$ on $BM$. Then notice that $QR$ is the perpendicular bisector of $BM$. Then $\Delta BQR\cong\Delta MQR$.

WLOG, $QC=1,BQ=x$. Then $MC=\frac{1+x}2,MQ=x$.

Then by Pythagoras on $\Delta MCQ$, we have $\left(\frac{1+x}2\right)^2+1=x^2$. Solving gives $x=\frac53$.

Thus, the crease divides $BC$ in a ratio of $5:3$.

$\boxed{\tiny Z}$

Let the side of the square be 2.

$\triangle BXY \sim \triangle BMC$. (Corrected. Originally was quoted as $\triangle BXY \sim \triangle AMC$.)

Setting up the corresponding ratios, we get $BX = \dfrac 54$ and hence $CX = 2 - BX = \dfrac 34$.

Extras (Completely re-written with the old picture replaced)

If, after folding, B falls on M, then the crease must be the perpendicular bisector of BM.

Let that perpendicular bisecor cut DC extended, BC, BM and AD at P, Q, R, S respectively.

By Pythagoras theorem, $MR = \dfrac {\sqrt 5}{2}$.

From $\triangle BQR \sim \triangle BMC$, we get $BQ = \dfrac {5}{4}$ and $QC = \dfrac {3}{4}$. In addition, QR is also known.

From $\triangle PCQ \sim \triangle PRM$, we get $PC = 1.5$.

From $\triangle PCQ \sim \triangle PDS$, we get $DS = \dfrac {7}{4}$.

Hence, $AS = 2 - DS = \dfrac 14$ and $DS : SA = 7 : 1$.


Why bother with Mathematics, if Gödel's Incompleteness Theorem is true? OK, maybe the title is exaggerated, but is it true that the rest of math is just "good enough", or a good approximation of absolute truth - like Newtonian physics compared to general relativity? How do we know that our "approximation" is the right one? Another analogy: Fundamental physics is also not  well-fundamented (where is the Higgs boson?) but most of the rest of the physics is on top of it, and it does its job well (it's a good-enough approximation).

Summary: According to the responses, math is indeed an imperfect domain, but can be seen as perfect for all practical purposes. In this case I wonder if math is indeed pure and identical across all possible universes. Maybe another universe comes up with a different set of axioms, more or less consistent than what we have now.

The first incompleteness theorem says that, under appropriate conditions, and for appropriate definitions of "true" and "provable," there exist statements that are true but not provable. This is interesting, but not a big deal. It is more or less a consequence of the existence of nonstandard models in first-order logic, which is an interesting feature of first-order logic, but mathematics is more than first-order logic.

The second incompleteness theorem says that, again under appropriate conditions, a sufficiently strong formal system can't prove its own consistency. Okay, this is kind of a big deal, but in practice it doesn't matter as much as it sounds like it does. Mathematicians don't actually do everything in a fixed formal system. There is a whole subject called reverse mathematics dedicated to finding the weakest formal systems that are capable of proving various things, which vary widely.

There is a possibility that ZFC, the formal system that sounds like it's what mathematicians work in (but isn't really), could be inconsistent. So what? If that ever happened we would just find another set of axioms to use. ZFC is a ridiculously strong formal system and in practice we never use its full strength, so it wouldn't really matter if it were inconsistent.

Mathematics is not the study of what formal statements can be proven in ZFC (although there are mathematicians who study this kind of thing). Proof: mathematics is thousands of years old, and ZFC is not.

Why do maths at all? It's fun, it's interesting, it tells us about the world we live in, and a bunch of other stuff. Gödel's incompleteness theorem doesn't tell us we're not really getting anywhere - it's a statement about the confines of pure logic, not a statement about how maths relates to the world. But even if it did tell us we were going to hit a wall somewhere, we've got pretty far along in our quest for knowledge so far - wouldn't you be interested in seeing how far we could get?


Problems with finding limit The function $f(x)$ has a derivative at the point $a$ and  $f(a) > 0$. I need to find the limit as n $\to + \infty$ of $$\left(\frac{f(a + \frac1n)}{f(a)} \right)^n$$ Substitution method?

You care about the limit
$$
\lim_{n\to \infty}\left(\frac{f(a+1/n)}{f(a)}\right )^n
$$
which is the same as
$$
\lim_{n\to \infty}
\exp\left(n\log\left(\frac{f(a+1/n)}{f(a)}\right )\right)\\
=\lim_{n\to \infty}\exp\left(\frac{\log\left(\frac{f(a+1/n)}{f(a)}\right )}{1/n}\right)
$$
Where we note that the expression in the exponential is the derivative of
the function
$$
g(x)=\log(f(x))
$$
at $a$ which by your assumption, exists. So the limit is in fact
$$
\exp\left(\frac{f'(a)}{f(a)}\right)
$$

We know there exists $\xi_n\in(a,a+1/n)$ such that $f(a+1/n)-f(a)=\frac{1}{n}f'(\xi_n)$. Now as $n\rightarrow\infty$, $f'(\xi_n)\rightarrow f'(a)$. So the limit becomes $$\lim_{n\rightarrow\infty}\left(\frac{f(a+1/n)}{f(a)}\right)^n=\lim_{n\rightarrow\infty}\left(1+\frac{f'(\xi_n)}{nf(a)}\right)^n=e^{\frac{f'(a)}{f(a)}}$$


Difference between Null set and empty set One of my friend asked this doubt.Even in lower class we use both as synonyms,he says that these two concepts have difference.Empty set $\{ \}$ is a set which does not contain any elements,while null set ,$\emptyset$ says about a set which does not contain any elements.

I could not make out that...is his argument correct ? if so how ?

In measure theory, a null set refers to a set of measure zero. For example, in the reals, $\mathbb R$ with its standard measure (Lebesgue measure), the set of rationals $\mathbb Q$ has measure $0$, so $\mathbb Q$ is a null set in $\mathbb R$. Actually, all finite and countably infinite subsets of $\mathbb R$ have measure $0$. In contrast, the empty set always refers to the unique set with no elements, denoted $\left\{ \right\}$, $\varnothing$ or $\emptyset$.

I would call "null sets" in the Measure Theory sense "sets of measure 0" just to avoid any confusion. In many parts of mathematics, "null set" and "empty set" are synonyms.


Are there more than one type of gradient? The lecturer is saying

  now the rise over run type gradient varies depending on where we choose our points

according to my knowledge rise over run is the gradient, they are the same. (may be I am wrong, please feel free to identify or fix the bugs)

So, are there more than one types of gradient?

Note: the key is on "type".

There is another type of gradient - and it's in the title of the video!

Derivative. The derivative is a different type of gradient that is NOT rise over run. But rather, a limit of rise over run.

Your instructor is probably saying that for a curve or surface the value of the "rise over run" gradient may be different at different places. The definition doesn't change.

(I haven't watched the video.)


$f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ twice Does there exist a continuous function $f: \mathbb{R} \to \mathbb{R}$ that takes each value in $\mathbb{R}$ exactly two times?

Suppose $f(a)=f(b)=0$.  Then on each of $(-\infty,a)$, $(a,b)$, $(b,\infty)$ the function $f$ is either positive or negative by the intermediate value theorem.  By continuity $f$ has either a max on $[a,b]$ which is strictly positive or a min which is strictly negative.  WLOG say it has a max which is positive.  The left and right intervals must have opposite signs or $f$ can't be surjective.  So say WLOG the left side is positive.  Then some (very small) positive value is achieved three times, once on the left interval and twice in the middle interval.

If possible let, f be such a continuous function in $\Bbb R$ which attains every value exactly twice.
Let $f(x_1)=f(x_2)=b$, with $x_1 \neq x_2$.
Then $f(x)\neq b$, for $x_1 \neq x_2$.
Then either $f(x)\gt b$ or $f(x)\lt b \forall x\in (x_1,x_2)$.
In case , when f(x)\gt b:
there is f(x_0)=max. {f(x):x\in (x_1,x_2)}.
Now we claim that, $f$ attains its maximum exactly once in $(x_1,x_2)$, otherwise $f$ will attain each value more than two times.
Let, $f$ attains its maximum exactly once at $c\in (x_1,x_2)$.
Again since $f$ attains its value exactly two times,  thus there exists $x_3$ ,outside of the interval $[x_1,x_2]$,
S.t. $f(x_3)=f(c)=d$ (,say) $\gt b$.
Then by IVP , we can conclude that f attains every value between $b$ and $d$ at least three times, CONTRADICTION.
Same argument can be applicable for $f(x)\lt b$.
Thus there doesn't exists any such continuous function.


Are the eigenvalues of $AB$ equal to the eigenvalues of $BA$? (Citation needed!) First of all, am I being crazy in thinking that if $\lambda$ is an eigenvalue of $AB$, where $A$ and $B$ are both $N \times N$ matrices (not necessarily invertible), then $\lambda$ is also an eigenvalue of $BA$?

If it's not true, then under what conditions is it true or not true?

If it is true, can anyone point me to a citation?  I couldn't find it in a quick perusal of Horn & Johnson.  I have seen a couple proofs that the characteristic polynomial of $AB$ is equal to the characteristic polynomial of $BA$, but none with any citations.

A trivial proof would be OK, but a citation is better.

If $v$ is an eigenvector of $AB$ for some nonzero $\lambda$, then $Bv\ne0$
and $$\lambda Bv=B(ABv)=(BA)Bv,$$ so $Bv$ is an eigenvector for $BA$ with the same eigenvalue. If $0$ is an eigenvalue of $AB$ then $0=\det(AB)=\det(A)\det(B)=\det(BA)$ so $0$ is also an eigenvalue of $BA$.

More generally, Jacobson's lemma in operator theory states that for any two bounded operators $A$ and $B$ acting on a Hilbert space $H$ (or more generally, for any two elements of a Banach algebra), the non-zero points of the spectrum of $AB$ coincide with those of the spectrum of $BA$.

Notice that $\lambda$ being an eigenvalue of $AB$ implies that $\det(AB-\lambda I)=0$ which implies that $$\det(A^{-1})\det(AB-\lambda I)\det(B^{-1})=0=\det(A^{-1}(AB-\lambda I)B^{-1})=\det((B-\lambda A^{-1})B^{-1}) $$ $$=\det(I-\lambda A^{-1}B^{-1}) = 0.$$
This further implies that $$\det(BA)\det(I-\lambda A^{-1}B^{-1})=\det(BA(I-\lambda A^{-1}B^{-1}))=\det(BA-\lambda I)=0,$$
i.e., $\lambda$ is an eigenvalue of $BA$.
This proof holds only for invertible matrices $A$ and $B$ though. For singular matrices you can show that $0$ is a common eigenvalue, but I can't think of a way to show that the rest of the eigenvalues are equal.


All smooth functions have Taylor expansion? The theories are:

$f(x)$ has Taylor expansion equals the remainder of Taylor polynomial converge to $0$.

a smooth function such as $f(x) = e^{-1/x^2}$, $x>0$, $f(x) = 0$, $x \leqslant 0$ do not has Taylor expansion near $0$.

However, I think about these propositions and find:
smooth - exist Taylor polynomial - select Peano remainder = $o(x^n$) - the remainder must converge to $0$ - smooth function must have Taylor expansion near $0$
I will be appreciated if someone can point out the error in this infer.

For every smooth function $f$, you can consider the Taylor series around $0$
$$\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$$
However, there are two possible obstructions in general to link directly this series to the function :

It is possible that the radius of convergence is $0$. Actually, one can prove that for every sequence $(a_n)$, there exists a smooth function $f$ satisfying $f^{(n)}(0)=a_n$. Therefore the Taylor series can be every series, so it can be a series which converges nowhere.

It is possible that the Taylor series converges everywhere, but is not equal to $f$ in a neighbourhood of $0$. The classical example is the one you mention with $e^{-x^2}$, you get a Taylor series which is the null series, and the function is non constant equal to $0$ in any neighbourhood of $0$.

Finally, you have the following condition for a function to be analytic over an interval $I$ : a smooth function $f : I \rightarrow \mathbb{R}$ is analytic over $I$ iff $\forall [a,b]\subset I$, there exists $M \in \mathbb{R}$ and $\alpha > 0$ such that $\forall n \in \mathbb{N}$, $$||f^{(n)}||_{\infty|[a,b]} \leq Mn!\alpha^n$$

Assume $f(x)$ as a smooth function and consider its $Maclaurin formula$,
then for every n:
$f(x) = \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k + o(x^n)$
select a sufficiently small neighborhood of zero, then:
$\lim\limits_{n\to+\infty}R_{n}(x) = \lim\limits_{n\to+\infty}(f(x) - \sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k)=\lim\limits_{n\to+\infty}o(x^n)=0$
then smooth function $f(x)=\lim\limits_{n\to+\infty}\sum_{k = 0}^n \frac{f^{(k)}(0)}{k!}x^k=\sum_{n \geq 0} \frac{f^{(n)}(0)}{n!}x^n$.
then smooth function $f(x)$ has Taylor expansion.
the above conclusion is wrong, because $e^{-1/x^2}$ is smooth and do not have Taylor expansion. However, what's the error during my infer?


proposition with implication EDIT: I first designated $x$, $y$ as irrational numbers. I mean rational.

I have this, In the question it says: For every $x$, $y$ being rational,there exists $z$ being rational so that: $x<z$ or $z<y$  Now, I have this: $\forall(x,y) \in\Bbb Q^2, \exists z\in\Bbb Q/(x<z)∨(z<y)$ Does this the signify the same as $\forall(x,y)\in\Bbb Q^2 \Rightarrow\exists z\in\Bbb Q/(x<z)\lor(z<y)$

The second statement you have written down is not a well-formed statement in first-order logic.

One way to see this is to look at any logic textbook where they inductively define what statements are admissible, and see that it's impossible to create something that looks like what you've written down.

Another way, though, is to think about the interpretation of $\rightarrow$. We say classically that $A\rightarrow B$ is true iff $B$ is true or $A$ is false and both $A$ and $B$ are statements. So for your statement, let's check whether the left hand side of the implication is true.

But the left hand side is just the fragment $\forall (x,y) \in \mathbb{Q}^2$, which is not a statement by itself.

EDIT: I apologize, there are conflicting conventions on this issue, relating to whether the variant of first order logic you're using has sorts. In some presentations $(\forall (x,y)\in \mathbb{Q}^2)A(x,y)$ is shorthand for $\forall x \forall y ((x,y)\in \mathbb{Q}^2 \wedge A(x,y))$.

Your first version says "For all rational ordered pairs (x, y), there exists a rational number z such that $x\lt z \lor z\lt y$."

Here is another way of stating what you seem to want to express, in the form of an implication:

$$\forall x, \forall y \Big((x\in \mathbb Q \land y \in \mathbb Q)\to \exists z\big(z\in \mathbb Q \land ((x\lt z)\lor (z\lt y)\big)\Big)\tag 1$$

$$\forall (x, y) \Big(\big (x, y) \in \mathbb Q^2)\to \exists z\big(z\in \mathbb Q \land ((x\lt z)\lor (z\lt y))\big)\Big)\tag 2$$

You've done most of the work, but in, example, $(1)$ above, we have (in loglish):  For all $x, y$, (if $x$ and $y$ are rational, then there exists a rational number $z$ such that $(x\lt z \lor z\lt y))$.

But you pretty much said the same thing, in your first proposal. My versions show the explicit implication operator; but your expression is another way to say the same thing.  In math, you'll often see statements written in your fashion.  It a more strict logic class, we use set membership as a predicate.


$\int_{- \infty}^{+ \infty} |f(t)| dt < \infty \implies \int_{-\infty}^{x} f(t) dt$ is continuous? I've found counter example for $(A),(D)$ and have shown except a bounded interval $F$ is uniformly continuous everywhere else. And so $(B)$ would imply $(C)$ is correct. But I can't show $(B)$ is correct. Please help someone.

HINT: For (B) consider $G_a(x)=\int_a^x f(t)\,dt$ and remind fundamental theorem of calculus.
$F(x)=\int_{-\infty}^a f(t)\,dt + G_a(x)$.


Is there an easy way to see associativity or non-associativity from an operation's table? Most properties of a single binary operation can be easily read of from the operation's table. For example, given
$$\begin{array}{c|ccccc}
\cdot & a & b & c & d & e\\\hline
a     & e & d & b & a & c\\
b     & d & c & e & b & a\\
c     & b & e & a & c & d\\
d     & a & b & c & d & e\\
e     & c & a & d & e & b
\end{array}$$
it is easy to check that it is closed (no elements occur in the table which don't occur as row or column index), commutative (the table is symmetric), has a neutral element (the row and column of $d$ are copies of the index row/column), and has an inverse element for each element (there's a $d$ in each row and column). In other words, almost all important properties can immediately be seen. The only part missing is associativity.

Therefore my question: Is there a simple way to see directly from the operation's table (i.e. without doing explicitly all the calculations) if an operation is associative?

Have you seen Light's associativity test?  According to Wikipedia, "Direct verification of the associativity of a binary operation specified by a Cayley table is cumbersome and tedious. Light's associativity test greatly simplifies the task."

If nothing else, the existence of Light's algorithm seems to rule out the possibility that anyone knows an easy way to do it just by looking at the original Cayley table.

Note also that, in general, one cannot do better than the obvious method of just checking all $n^3$ identities of the form $(a\ast b)\ast c = a\ast (b\ast c)$.  This is because it is possible that the operation could be completely associative except for one bad triple $\langle a,b,c\rangle$. So any method that purports to do better than this must only be able to do so in limited circumstances.

A set with an operation used to build a four-dimensional multiplication table with entries $X_{abcd} = (A(BC))D$ is associativ iff $X_{abc1} = X_{1abc}$, with 1 being unity. So it would be a certain three-dimensional symmetry of a four-dimensional table.


There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number I just took an olympiad and I'm wondering how this problem is solved.

Problem: There exists a power of 2 such that the last five digits are all 3's or 6's. Find the last 5 digits of this number.

Please don't find the solution on your computer and then work backward because you might not be able to fully explain the insight that a person working without a computer would need.

66336:

The last digit is 6 because 3 is not divisible  by 2.

The second-to-last digit is 3 because 66 is not divisible  by 4.

The third-to-last digit is 3 because 636 is not divisible by 8.

The fourth-to-last digit is 6 because 3336 is not divisible by 16.

The fifth-to-last digit is 6 because 36336 is not divisible by 32.

(Edit: This is sufficient because $2^{n}$ divides $10^{n}$, so it doesn't matter what the preceding digits are.)

Another easy way would be to work with binary representation.

The number in question is a power of 2 and all powers of a number n when represented in the same base n must have their most significant digit as 1 and all other digits as 0.

e.g. 2 in binary is 10, 4 is 100, 8 is 1000...., 1024 is 10,00,00,00,000 and so on. Your number will also look like this.

But, of course, this method is equivalent to sequentially checking divisibility by 2, 4, 8, 16 and 32.


Equation Involving Exponential Functions of Differing Bases I'm completely stumped here.
$$
2^{x+1} = 5^x
$$
If someone could explain how to solve for $x$ I'd be grateful.

Thanks.

Here's the step by step solution

$$2^{x+1} = 5^x$$

using the law of indices $$a^{b+c} =a^b a^c$$

$$log[2^x2^1] = log[5^x]$$

Taking log base 10 on both the sides and using one of the properties you get,

$$log (2^x) + log(2^1) = log(5^x)$$

using yet another property of log's we get

$$xlog2 + 1log2 = xlog5$$

x(0.3010) + 1(0.3010) = x(0.6989)

Now you know how to solve for x right?


The sequence $\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$ is bounded above . A sequence $(\phi_n)$ is defined as follows : $$\phi_n= 1+\frac{1}{2^4}+\frac{1}{3^4}+\ldots+\frac{1}{n^4}$$
Show that the sequence is convergent.

Because this sequence is monotonic, proving it is bounded above will be sufficient to prove that it is convergent.

So how to show that this sequence is bounded above?

Here's a simple approach :

$$k^4 \geq k^2 \geq k(k-1)$$ so :

$$\phi_n=1+\sum_{k=2}^{n} \frac{1}{k^4} \leq 1+\sum_{k=2}^{n} \frac {1}{k(k-1)}$$ but the later sum telescopes because $$\frac{1}{k(k-1)}=\frac{1}{k-1}-\frac{1}{k}$$ so :

$$\phi_n \leq 1+1-\frac{1}{n}=2-\frac{1}{n}<2$$ so this sequence is bounded and then because of monotony it's also convergent .

As an aside the sum actually converges to :$$\zeta(4)=\frac{\pi^4}{90}$$


Reducing primes in $\mathbb Z[i]$ Let $4k+1=p$ be a prime.

Assume you do not know $p=a^2+b^2$(this is what we intend to prove).

Over $\mathbb Z[i]$, how does one prove that $p$ splits into conjugates?

That is, if $p = (a+bi)(c+di)$, then $a+bi=c-di$?

We know that $a-bi=t(c+di)$ where $t\in \mathbb R$, could give: $p=t(a^2+b^2)$. It could be that $t=p/m^2$ where $m^2=a^2+b^2$. How do you show $t=\pm 1$?

Does this work?

$a-bi=t(c+di)\implies (a-bi)(a+bi)=t(c+di)t(c-di)\implies(a^2+b^2)=t^2(c^2+d^2)\implies t^2=1(\mbox{since }a^2+b^2=c^2+d^2=p)\implies t=\pm 1.$

The ring $\mathbf{Z} / p \mathbf{Z}$ is a field with $p-1$ nonzero elements. Since $4 | p-1$, there is a multiplicative subgroup of order $4$.

A field can only have two square roots of unity, therefore $\mathbf{Z} / p \mathbf{Z}$ already has a fourth root of unity $m$.

Pick your favorite reason why this implies $p\mathbf{Z}[i]$ cannot be a prime ideal of $\mathbf{Z}[i]$: e.g. some simple things things you can use are

$\mathbf{Z}[i]/p\mathbf{Z}[i] \cong (\mathbf{Z}/p\mathbf{Z})[i]$ is not a field, because it has too many fourth roots of unity
$i-m \not\in p\mathbf{Z}[i]$
$p | (m-i)(m+i)$ but neither $p | (m-i)$ nor $p | (m+i)$

The ring $\mathbf{Z}[i]$ is a Euclidean domain, which implies it is a unique factorization domain.

Since the prime factorization of $p$ cannot simply be $p$ (nor can it be $(-1) \cdot (-p)$) and cannot be any other product of integers, $p$ must have a prime factor of the form $a+bi$. By conjugation (which is an automorphism of $\mathbf{Z}[i]$ that is the identity on $\mathbf{Z}$), $a-bi$ is also a prime factor of $p$.

Thus $(a+bi)(a-bi) = a^2 + b^2$ is a divisor of $p$. Since it's an integer bigger than $1$, we must actually have $a^2 + b^2 = p$.

Inspired by Will Jagy's comment.

We know that $(2k)!^2\equiv(4k)!\equiv(p-1)!\equiv-1\mod p$ and so $x^2+1\equiv 0 \mod p$ has a solution for $x$. Hence $p|(x^2+1)\implies \frac{(x+i)(x-i)}{p}\in\mathbb Z\implies p=(a+bi)(c+di)$.

We know that $a-bi=t(c+di)$ where $t\in \mathbb R$.

We know that $a^2+b^2=c^2+d^2=p$ as $p^2=p\cdot \bar{p} = (a+bi)(a-bi)(c+di)(c-di)=(a^2+b^2)(c^2+d^2)\implies p=a^2+b^2$ and $c^2+d^2.$

$\mbox{So }a-bi=t(c+di)\implies (a-bi)(a+bi)=t(c+di)t(c-di)\implies(a^2+b^2)=t^2(c^2+d^2)\implies pt^2=p\implies t^2=1\implies t=\pm 1\implies p=\pm(a+bi)(a-bi).$

Hence $p=a^2+b^2$ since $p>0$.


Prove that 360 divides (a-2)(a-1)a.a.(a+1)(a+2) there's a question which asks to prove that

360 | a2(a2-1)(a2-4)

I attempted it in the following manner.

a2(a2-1)(a2-4) = (a-2)(a-1)(a)(a+1)(a+2)(a)

The first 5 terms represent the product of 5 consecutive terms. Hence,
One of them will have a factor of 5
One of them will have a factor of 4
One of them will have a factor of 3
One of them will have a factor of 2
=> 5 x 4 x 3 x 2 = 120 will divide the given expression for sure.

Now, how do i bring a factor of 3. I know that i have missed one power of a. But, i think i am missing something. Help will be appreciated

Thank you

Hint: both $(a-2)(a-1)a$ and $a(a+1)(a+2)$ are products of three consecutive integers.


Simplex Method -why does Gauss Elimination produce a new better solution? In the simplex method, a tableau is created. An entering variable is generated and then using a criteria the exiting variable is also selected. This gives a pivot column and row and pivot element.

The next step in the basic simplex algorithm is to perform Gaussian Elimination. For example
starting with this tableau

through the Gaussian process this tableau is generated

What is the reasoning behind using Gaussian Elimination? Why does it generate a new and a better solution (a better z value)?
Thank you.

Just take a look here
https://github.com/alvarogatenorio/Investigacion-Operativa

The link contains a pdf of operations research notes that I'm writing right now. The first section of chapter 2 contains all you need. However, if you are not familiar with the simplex theoretical foundations you will need to read the first chapter as well.

Unfortunatelly, everything is written in spanish, but I think is pretty understandable with a little help of google translate.


Does $\lim\limits_{x\rightarrow \infty}f'(x)$ exist? Let $f: \Bbb R \rightarrow \Bbb R$ be differentiable with $f'$ uniformly continuous. Suppose $\lim\limits_{x\rightarrow \infty}f(x)=L$ for some $L$. Does $\lim\limits_{x\rightarrow \infty}f'(x)$ exist?

I have no idea about this problem. Could you please give a hint? Thank you very much for your help.

Outline:

Without loss of generality, $L=0$ (if not, consider $g=f-L$).

Suppose by contradiction that $f'(x)\not\xrightarrow[n\to\infty]{}0$. Negating the definition, it means there exists $\varepsilon_0>0$ and a sequence $(x_n)_{n\geq 0}$ with $x_n\xrightarrow[n\to\infty]{} \infty$ such that $\lvert f'(x_n) \rvert > \varepsilon_0$ for all $n\geq 0$. By a standard argument (pick, out of the two, the subsequence with the most points; then wlog you can assume that the derivative is positive, otherwise consider $-f$ instead of $f$) we get a sequence $(y_n)_{n\geq 0}$ with $y_n\xrightarrow[n\to\infty]{} \infty$ such that $f'(y_n) > \varepsilon_0$

Let $\delta > 0$ be the uniform continuity modulus for $f'$ (which is assumed uniformly continuous) and $\varepsilon\stackrel{\rm def}{=}\frac{\varepsilon_0}{2}$. We get that for any $n\geq 0$, and any $x\in[y_n-\delta, y_n+\delta]$, we have $f'(x) \geq \varepsilon_0-\varepsilon = \frac{\varepsilon_0}{2}$. But then, you now something about $f$ on $[y_n-\delta, y_n+\delta]$: namely, that it "goes away" from $y_n$ at least at the rate of an affine function with  slope $\frac{\varepsilon_0}{2}$.

Now use that to show that you must have infinitely many points $z_n$ (one around each $y_n$) such that $\lvert f(z_n) \rvert \geq \frac{\delta\varepsilon_0}{2}>0$. How does that contradict the assumption that $f(x)\xrightarrow[x\to\infty]{} 0$?

Hint: If the limit on f exists, it means f becomes constant as x goes to infinity. What can you say about the derivative of a constant function?


Why is it so important for the characteristic value of the field of a lie algebra to not be two for many propositions? In reading my Lie algebra text, I see a lot of propositions starting with, "If char F does not equal 2, then..." For example, if char F does not equal 2, then o(n,F) is a subalgebra of sl(n,F).

I am failing to see the importance in most cases. Could someone give me an example when char F = 2 will be the downfall? I was thinking about fields where x = -x but couldn't pinpoint the error. I'm sure it's a foolish miss on my part.

Thanks sincerely!

Characteristic three is also problematic for Lie algebras. The appearence of the two characteristics is related to the fact that the identities that define Lie algebras —antisymmetry and the Jacobi identity— are related to the symmetric groups of degree $2$ and $3$, and these have very simple representation theory except when the characteristic is $2$ or $3$.

You hit the nail on the head with thinking "x = -x" because that is equivalent to 2x=0 for all x. I admit openly that I know little about Lie Algebras but I do understand the difficulty of char F = 2.

If char F = 2 the the field F is isomorphic to {0,1} so anytime you appeal to F all it offers is a 0 or a 1 and in that case it is probable your exercise will be rather trivial.

If you are interested in Galois Theory look up Keith Conrad's papers about finding Galois groups via discriminants. He always sets out with char <> 2.


How to Solve Problem Similar to IMO(1995) Problem Two years ago conjecture,Today I again remind of the problem

(conjecture) Let $ n$ be an postive integer number. How many $ n$-element subsets $A$ of $ \{1,2,\dots,2n\}$ are there such that $1+2+\cdots+2n$ is divisible by the sum of the elements of $A$.

I have found this similar problem:IMO (1995) last problem:IMO
two year ago idea:
let $$A=\{x_{1},x_{2},\cdots,x_{n}\},1\le x_{1}<x_{2}<\cdots<x_{n}\le 2n$$
and we have
$$1+2+3+\cdots+2n\equiv 0\pmod{x_{1}+x_{2}+\cdots+x_{n}}$$
Today ADD it,  By Now I have read Problem and from the book Chapter 7 example 4 (simaler problem):
(Solved )  Let $f(n)$ be the number of subsets of  $1,2,3,\cdots,n$ whose elements sum  to 0 $\pmod n$,the empty set is included,having the element sum equal to zero,Prove that
$$f(n)=\dfrac{1}{n}\sum_{d|n,d~odd}\varphi(d)2^{\frac{n}{d}}$$
Solution: Let
$$g(X)=\prod_{i=1}^{n}(1+X^i)=\sum_{k\ge 0}a_{k}X^k$$
if let $\varepsilon =e^{\frac{2i\pi}{n} }$,it is clear
$$f(n)=\sum_{j\ge 0}a_{jn}$$
and other hand easily be computed
$$\sum_{j=1}^{n}g(\varepsilon ^j)=n\sum_{j\ge 0}a_{jn}$$
if let $d=\dfrac{n}{\gcd{(n,j)}}$,and use well known
$$x^d-1=(x-\varepsilon^j)(x-\varepsilon^{2j})\cdot (x-\varepsilon^{dj})$$
so we have
$$(1+\varepsilon^j)(1+\varepsilon^{2j})\cdot (1+\varepsilon^{dj})=2$$
this shows that
$$g(\varepsilon ^j)=2^{n/d},d ~~is odd$$
and $0$ otherwise.
so
$$\dfrac{1}{n}\sum_{j=1}^{n}g(\varepsilon ^j)=\dfrac{1}{n}\sum_{d|n,d~odd}\varphi(d)2^{\frac{n}{d}}$$
if $n=p$ be prime,then it's 1995 IMO problem.
But for conjecture Now I don't solve it by now!

Hint:  Note that $\sum_{i=1}^{2n}i = n(2n+1)$  The sum of an $n$ element subset of $A$ is at least $\frac 12n(n+1)$, so the only sums that can divide $n(2n+1)$ are  $\frac 13n(2n+1)$, $\frac 12n(2n+1)$ and $n(2n+1)$.  You need to find how many $n$ element subsets sum to one of these.  The first and third may not be a whole number, and should be ignored if so.

$\frac {n(2n+1)}{\frac {n(n+1)}{2}}<4$ ,so  equation which a sum divided is 2 or 3.
(i)When sum of two elements is 2n+1,  $$2n+1=(1,2n)(2,2n-1)\ldots(n,n+1)$$n pairs.By choosing $\frac n2$, $_nC_{\frac n2}$ pairs.This sum is just half of equation.

(ii)When sum of two elements is n or 3n$$(1,n-1)(2, n-2)\dots(\frac n2-1,\frac n2+1)$$
$$(n,2n)(n+1,2n-1)\ldots(\frac {3n}2-1,\frac {3n}2+1)$$This sum are $_\frac {n-2}2C_k$,$_\frac {n-2}2C_\frac {n-2k}2$ by choosing k,$\frac {n-2k}2$ .
These sum is$$\tag{1}(n*k)(3n*(\frac n2-k))$$
By the way,there is no common diviser of $(n,2n+1)$ and (1) is not diviser of equation whatever k is.

(iii)When sum is 3,
$(2n-2,1,2)(2n-6,3,4)\ldots(\frac {2n}3+1,\frac {2n}3-1,\frac {2n}3)$

$\frac {n}3+1$ pairs,$_{\frac n3+1}C_\frac n3=\frac n3+1 $

So answers are,using $_{\frac n3}C_{\frac n3}=1$

if $n=6k-2,6k-4,$ $_nC_{\frac n2}+1$

if $n=6k-1,6k-5$, $_nC_{\frac {n-1}2}+1$

if $n=6k$, $_nC_{\frac n2}+{\frac n3}+1 $

if $n=6k-3$,$_nC_{\frac {n-1}2}+{\frac n3}+1$


Prove that H is orthocentre of ABC using inversion Three equal circles pass through a given point $H$ and meet one another two by two at $A,B,C$ prove that $H$ is orthocentre of triangle $ABC$.

My try -

I proved it using elementary geometry methods quite easily but i also want to prove it using inversion...

so first i invert about H to make all three circles lines but not able to go further to prove that it is orthocentre of ABC...

Any hints?

Invert at $H$. The fact that the three circles through $H$ are equal means that, if $A',B',C'$ are the images of $A,B,C$, then the distances from $H$ to $A'B'$, $B'C'$, and $C'A'$ are the same, because the point diametrically opposite $H$ on $(HAB)$, for example, inverts to the foot from $H$ to $A'B'$. Therefore $H$ is the incenter of $A'B'C'$. Now, let $X,Y,Z$ be the $A'$-, $B'$-, and $C'$-excenters of $A'B'C'$. We see that $H$ is the orthocenter of $XYZ$ and that a negative inversion about $H$ sends $A'\to X$, etc., so $ABC$ is directly similar at $H$ to $XYZ$, and so $H$ is the orthocenter of $ABC$.

Observe an inversion with center at $C$ and radius $r= CH$.

Let $A\mapsto A'$ and $B\mapsto B'$.

Since circles (Y) and (Z) have equal radius and go through $C$ they map to a lines $CA'$ amd $CB'$ which are at equal distance from $C$. So $HC$ is angle bisector for $\angle A'HB'$. Now we have $$\angle HBC = \angle B'HC = \angle A'HC = \angle HAC =:\alpha $$

By simmetry we have $$\angle HBA = \angle HCA =:\beta $$ and $$\angle HAB = \angle HCB =:\gamma $$

Now it is clealry $$\alpha +\beta +\gamma = 90^{\circ}$$ so we are done.


Help me prove: sin(A+B) = sinA cosB + cosA sinB Can you help me prove that: sin(A+B) = sinA cosB + cosA sinB?

Thanks!

Here are  two for the price of one (using Euler's formula):

$$\color{red}{\cos(A+B)}+i\color{green}{\sin(A+B)} \equiv e^{i(A+B)} \equiv e^{iA} \times e^{iB}$$
$$\equiv [\cos(A)+i\sin(A)][\cos(B)+i\sin(B)]$$
$$\equiv \color{red}{[\cos(A)\cos(B)-\sin(A)\sin(B)]}+i\color{green}{[\sin(A)\cos(B)+\cos(A)\sin(B)]}$$

Now equate imaginary parts to give the result for $\sin(A+B)$ (and, if you want, equate real parts to give the result for $\cos(A+B)$).

May be a little unconventional, but assuming we have an isosceles right angled triangle.

Then, $sin 90^\circ=1$

$$=\frac{1}{2}+\frac{1}{2}$$

$$=\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}+\frac{1}{\sqrt2}\cdot\frac{1}{\sqrt2}$$

$$=cos 45^\circ \cdot sin 45^\circ+sin 45^\circ \cdot cos 45^\circ$$

The similar can be proved for a scalene triangle as well.


Show that $\lim_{n \to \infty}\frac{n^2}{n!}=0$ Show that $\lim_{n \to \infty}\frac{n^2}{n!}=0$. Can you give me a hint? Thanks to all.

Not the most elegant but...

Hint: Use Stirling's Approxmation:

$$\lim \frac{n^2}{n!}$$

$$ = \lim \frac{n^2}{n!} \frac{n!}{\sqrt{2 \pi n} (\frac{n}{e})^n}$$

$$ = \lim \frac{n^2}{\sqrt{2 \pi n} (\frac{n}{e})^n}$$

Now compute:

$$\lim \frac{n^2}{\sqrt{n} (\frac{n}{e})^n}$$

$$ = \lim \frac{n^{3/2}}{(\frac{n}{e})^n}$$

Use LHR somewhere


Can you calculate $\frac{dx}{dy}$ by finding $\frac{dy}{dx}$ and flipping it over? This may be a silly question but for example if you had the gradient at $x=4$ of $y=x^2+1$, then can you just calculate $\frac{dx}{dy}$ by finding $\frac{dy}{dx}$ and flipping it over? Or must you make $x$ the subject and differentiate?

If $f$ is invertible and differentiable, with $f'$ never zero, then $f^{-1}$ is also differentiable and

$$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}. $$

Does this help?

Yes, you can. If both derivatives exist, they are reciprocal, that is

$\frac{dy}{dx}\frac{dx}{dy}=1$


Finding a basis for a field extension $\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q}(\sqrt{2}).$ I am looking to find a basis for the field extension $$\mathbb{Q}(i,\sqrt[4]{2})/\mathbb{Q}(\sqrt{2}).$$

Clearly as far as I can tell the elements $i$ and $\sqrt[4]{2}$ would be in the basis, as neither of these elements is in the field $\mathbb{Q}(\sqrt{2})$. However, when it comes to finding other elements in the basis, all that I could really think of at first was $i\sqrt[4]{2}$. Then I supposed that we would also need $\sqrt[4]{2}^3$ and $i\sqrt[4]{2}^3$ in the basis too, and later wondered whether $i \sqrt{2}$ would also be required. I think not because it would be covered under the element $i$ as $\sqrt{2}$ is contained in the field but I am not certain.

As it took me a while to find each of these 5 elements, I am wondering whether this is actually all the elements that would be required to form a basis, whether I am missing any elements or have too many. Any help would be appreciated thanks :)

Note that $i\notin\mathbb{Q}(\sqrt[4]{2})$, so $\mathbb{Q}(i,\sqrt[4]{2})$ has degree $8$ over $\mathbb{Q}$. Hence it has degree $4$ over $\mathbb{Q}(\sqrt{2})$.

A basis for $\mathbb{Q}(\sqrt[4]{2})$ over $\mathbb{Q}(\sqrt{2})$ is $\{1,\sqrt[4]{2})$.

Can you finish? You still have to add $i$.

I think (2^¼)^2=√2 this imply that √2∈Q(2^¼) that is Q(√2) Contained in Q(2^¼) therefore degree of i over Q(√2) is 2 that is  degree of Q(2^¼, i) over Q(√2) is 2 therefore basis of Q(2^¼, i) over Q(√2) contains only two element namely {1,i}.


The Jezebel probability I came across this brain teaser that has two parts.

The first part is already answered here: In a family with two children, what are the chances, if one of the children is a girl, that both children are girls?

The second question is: In a family with two children, what are the chances, if one of the children is a girl called Jezebel, that both children are girls?

SPOILER

The answer is supposedly cca (source [czech])
$(2*2*4*8)/(4*4*4*4)$

There is an explanation, but it sounds like troll logic.

The first part is straightforward, the possibilities are $BG, GB, GG$, hence the probability of both being girls is ${1 \over 3}$.

The second part is also straightforward, assuming that both girls are not named Jezebel. This time let $J$ represent Jezebel and $G$ represent a girl not named Jezebel. Then the possibilities are $BJ, JB, JG, GJ$, hence the probability of both being girls is ${1 \over 2}$.

Well, there are 9 possibilities, where G denotes a girl not called Jezebel:

BB
BG
GB
GG
BJ
JB
GJ
JG
JJ

What are the probabilities of each event? Let's suppose the probability of the parents naming a girl child Jezebel is $p$ if they haven't already got a child called Jezebel, and $q$ if they have. Then the five relevant probabilities are:

BJ : $p/4$

JB : $p/4$

GJ : $(1-p)p/4$

JG : $p(1-q)/4$

JJ : $pq/4$

The sum of these is $p - p^2/4 = p(1-(p/4))$, so the conditional probabilities are:

BJ given $\exists$J : $1/(4-p)$

JB given $\exists$J : $1/(4-p)$

So the probability of them both being girls is $1-(2/(4-p))$ = $(2-p)/(4-p)$. So if $p$ is very small, then the probability is almost $1/2$. If $p$ is very large (i.e. nearly $1$),  then the probability is just over $1/3$. (This is the "one of them is a girl" case).

Since Jezebel is not in the top 1000 girls' names in the US (source: http://www.thinkbabynames.com/meaning/0/Jezebel), $p$ is very small and the probability of them both being girls is almost $1/2$.


Solving for $(x,y): 2+\frac1{x+\frac1{y+\frac15}}=\frac{478}{221}$ Solving for $x,y\in\mathbb{N}$:

$$2+\dfrac1{x+\dfrac1{y+\dfrac15}}=\frac{478}{221}$$

This doesn't make any sense; I made $y+\frac15=\frac{5y+1}5$, and so on, but it turns out to be a very complicated fraction on the left hand side, and I don't even know if what I got is correct.

Is there a more mathematical way to approach this problem?

Yes, there is a very large and important mathematical theory, called the theory of Continued Fractions. These have many uses, both in number theory and in analysis (approximation of functions).

Let's go backwards.

The number $\frac{478}{221}$ is $2+\frac{36}{221}$, which is $2+\frac{1}{\frac{221}{36}}$.

But $\frac{221}{36}=6+\frac{5}{36}$, which is $6+\frac{1}{\frac{36}{5}}$.

Finally, $\frac{36}{5}=7+\frac{1}{5}$.

Now compare the results of these calculations with your expression.

To see uniqueness,

we look at Danny's solution.

Let's see,
\begin{align}
\frac{478}{221}&=2+\frac{36}{221} &\text{2 is the only integer which will make the leftover fraction less than 1}\\
&=2+\dfrac1{\frac{221}{36}}\\&
=2+\frac1{6+\frac5{36}} &\text{Here, 6 is the only number which will make the left over fraction positive and less than 1}\\
&=2+\dfrac1{6+\dfrac1{\frac{36}{5}}}\\
&=2+\dfrac1{6+\dfrac1{7+\frac15}} &\text{and here, 7 is the only number}
\end{align}

Thus, $x=6,y=7$.


What does it mean for a random variable to "admit" a distribution? Can someone explain the word "admit" and explain what would happen if it does not admit a distribution?

Not every probability distribution on the real numbers can be described by a probability density function: for example, the probability distribution given by (on the variable $X$)

$$ P(X = a) = \begin{cases} 1/2 & a=0,1 \\ 0 & \text{otherwise} \end{cases} $$

cannot be expressed by a probability density function.

"Admit" is sometimes used to express the notion that you're allowed or able to do something -- in this case, saying that a distribution "admits a density function" means that the distribution the phrase refers to really does have a probability density function.

A random variable, $X$, is a measurable function from a probability space $(\Omega,\mathcal{F},\mathbb{P})$ to $(\mathbb{R},\mathcal{B}_{\mathbb{R}})$. The induced measure on $\mathcal{B}_{\mathbb{R}}$, defined by $$\mu(A)=\mathbb{P}(X^{-1}(A))$$ is the distribution of $X$.


Solve the One Dimensional Eigenvalue Problem: $u_{tt} - u_{xx} = m^2 u$ Solve the One Dimensional Eigenvalue Problem:
  $$u_{tt} - u_{xx} = m^2 u$$

I have learned how to solve PDEs of the form $u_{tt} - u_{xx} = f(x,t)$ using d'Alembert's solution. However, I have not yet studied PDEs of this form. How does one solve this PDE?

Suppose we try separation of variables, setting

$u(x, t) = f(t)g(x); \tag 1$

then

$u_{xx} = f(t) g''(x), \tag 2$

$u_{tt} = \ddot f(t) g(x), \tag 3$

whence the equation

$u_{tt} - u_{xx} = m^2 u \tag 4$

becomes

$\ddot f g - f g'' = m^2 fg; \tag 5$

if we assume for the moment that $f(t) \ne 0 \ne g(x)$, then we may divide (5) through by $fg$, yielding

$\dfrac{\ddot f}{f} - \dfrac{g''}{g} = m^2, \tag 6$

or

$\dfrac{\ddot f}{f} - m^2 = \dfrac{g''}{g}. \tag 7$

We can now argue in the typical manner that, since the left-hand side depends only on $t$ but the right-hand side only on $x$, they must each equal the same constant value which, for reasons which will hopefully soon become clear, we denote by $-\lambda^2$:

$\dfrac{\ddot f}{f} - m^2 = -\lambda^2 = \dfrac{g''}{g}, \tag 8$

which leads to

$\ddot f + (\lambda^2 - m^2) f = 0, \tag 9$

and

$g'' + \lambda^2 g = 0.\tag{10}$

(10) has solutions of the general form

$g_\lambda(x) = a_+(\lambda)e^{i\lambda x} + a_-(\lambda) e^{-i\lambda x}, \tag{11}$

where $\lambda$ and the $a_\pm(\lambda) \in \Bbb C$ are determined by boundary/initial conditions placed upon $g(x)$; we can't get much more specific at this point without such information.  For each admissible $\lambda$, we then have a solution for $f(t)$:

$f_\lambda(t)= b_+(\lambda) e^{i\sqrt{\lambda^2 - m^2}t} + b_-(\lambda)e^{-i\sqrt{\lambda^2 - m^2}t}; \tag{12}$

therefore,

$u_\lambda(x, t) = f_\lambda(t)g_\lambda(x)$
$= ( b_+(\lambda) e^{i\sqrt{\lambda^2 - m^2}t} + b_-(\lambda) e^{-i\sqrt{\lambda^2 - m^2}t})(a_+(\lambda)e^{i\lambda x} + a_-(\lambda) e^{-i\lambda x}); \tag{13}$

it is easy to check that

$u_{\lambda tt} - u_{\lambda xx} = m^2 u_\lambda. \tag{14}$

Solutions of the form (13) for different $\lambda$ may be composed by summation or integration over $\lambda$ to yield more complicated solutions.

Finally, it is of course worth observing that the temporal frequencies $\sqrt{\lambda^2 - m^2}$ may become imaginary if $\lambda^2 < m^2$; the we obtain temporally exponentially growing/decreasing solutions $u(x, t)$.

In the above, I have decided to associate the $m^2$ terms with $f(t)$, but we could just as well have taken

$g'' + (\lambda^2 + m^2)g = 0 \tag{15}$

and obtained a different, though similar, set of solutions to (4).  In fact, if we set

$m^2 = m_t^2 - m_x^2, \tag{16}$

we can write

$\dfrac{\ddot f}{f} - \dfrac{g''}{g} = m_t^2 - m_x^2, \tag{17}$

whence

$\dfrac{\ddot f}{f} - m_t^2 = -\lambda^2 = \dfrac{g''}{g} - m_x^2, \tag{18}$

which leads to a more symmetrical analysis and set of solutions.

This could go on forever, but I can't, so I'll leave it at that.  I would like to say in closing, however, that all this stuff smells like the Special Theory of Relativity and Particle Physics to my somewhat knowlewdge-able knose.

Let $p=mx$ ,

Then $u_{tt}=m^2u_{pp}+m^2u$

Let $q=mt$ ,

Then $u_{qq}=u_{pp}+u$

Similar to Solve initial value problem for $u_{tt} - u_{xx} - u = 0$ using characteristics,

Consider $u(p,a)=f(p)$ and $u_q(p,a)=g(p)$ ,

Let $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^n}{n!}\dfrac{\partial^nu(p,a)}{\partial q^n}$ ,

Then $u(p,q)=\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n}}{(2n)!}\dfrac{\partial^{2n}u(p,a)}{\partial q^{2n}}+\sum\limits_{n=0}^\infty\dfrac{(q-a)^{2n+1}}{(2n+1)!}\dfrac{\partial^{2n+1}u(p,a)}{\partial q^{2n+1}}$

$u_{qqqq}=u_{ppqq}+u_{qq}=u_{pppp}+u_{pp}+u_{pp}+u=u_{pppp}+2u_{pp}+u$

$u_{qqqqqq}=u_{ppppqq}+2u_{ppqq}+u_{qq}=u_{pppppp}+u_{pppp}+2u_{pppp}+2u_{pp}+u_{pp}+u=u_{pppppp}+3u_{pppp}+3u_{pp}+u$

Similarly, $\dfrac{\partial^{2n}u}{\partial q^{2n}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k}u}{\partial p^{2k}}$

$u_{qqq}=u_{ppq}+u_q$

$u_{qqqqq}=u_{ppqqq}+u_{qqq}=u_{ppppq}+u_{ppq}+u_{ppq}+u_q=u_{ppppq}+2u_{ppq}+u_q$

$u_{qqqqqqq}=u_{ppppqqq}+2u_{ppqqq}+u_{qqq}=u_{ppppppq}+u_{ppppq}+2u_{ppppq}+2u_{ppq}+u_{ppq}+u_q=u_{ppppppq}+3u_{ppppq}+3u_{ppq}+u_q$

Similarly, $\dfrac{\partial^{2n+1}u}{\partial q^{2n+1}}=\sum\limits_{k=0}^nC_k^n\dfrac{\partial^{2k+1}u}{\partial p^{2k}\partial q}$

$\therefore u(p,q)=\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^nf^{(2k)}(p)(q-a)^{2n}}{(2n)!}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{C_k^ng^{(2k)}(p)(q-a)^{2n+1}}{(2n+1)!}$


Upper bound for $\gcd(a,b)$ if $\frac{a+1}{b}+\frac{b+1}{a}\in\Bbb{N}$ Suppose that $a,b$ are two positive integers so that $\frac{a+1}{b}+\frac{b+1}{a}$ is also a positive integer.Find the best upper bound for $\gcd(a,b)$.

My work:
$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a(a+1)+b(b+1)}{ab} \in \Bbb{N} \implies ab|a^2+b^2+a+b , 2ab \implies ab|(a+b)(a+b+1)\implies ab|a+b\ or\ ab|a+b+1$
Now may I infer that $a=b=1\ or\ a=b=2$? If yes how may we set an upper bound for $\gcd(a,b)$??!!

For the equation. $$\frac{m+1}{n}+\frac{n+1}{m}=a$$

If are solutions of the equation Pell.  $p^2-(a^2-4)s^2=1$

Then the formula is:

$$n=2(p-(a+2)s)s$$

$$m=-2(p+(a+2)s)s$$

And another solution:

$$n=\frac{2p(p+(a-2)s)}{a-2}$$

$$m=\frac{2p(p-(a-2)s)}{a-2}$$

If are solutions of the equation Pell. $p^2-(a^2-4)s^2=4$

Then the formula is:

$$n=\frac{p-(a-2)s+2}{2(a-2)}$$

$$m=\frac{p+(a-2)s+2}{2(a-2)}$$

You can write a more General formula.  http://www.artofproblemsolving.com/community/c3046h1046841___

Fortunately with the help of one of my friends I got to the right answer as follows:
Assuming $d=\gcd(a,b),a=a'd,b=b'd$ we have: $$\frac{a+1}{b}+\frac{b+1}{a}=\frac{a'^2d+b'^2d+a'+b'}{a'b'd}$$
Now  it's required that $d|a'+b'$ so that the expression is an integer , then:
$$d^2|a'd+b'd\implies d^2|a+b \implies d^2\leqslant a+b \implies \color{red}{d\leqslant \sqrt{a+b}}$$
Tha's all.


How to calculate $f(n)=1\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}.....n\binom{n}{n}$ $$f(n)=1\binom{n}{1}+2\binom{n}{2}+3\binom{n}{3}.....n\binom{n}{n}$$

Is there a formula for this?

Hint: Expand $n(1+x)^{n-1}$ using the binomial theorem, then put $x=1$.


Norm of a left shift operator Left shift operator is $L:\ell^2\to\ell^2$ defined by $$(x_1,x_2,x_3,x_4,\ldots)\mapsto (x_2,x_3,x_4,\ldots).$$

This is not an isometry apparently, so $\|Lx\|\ne \|x\|$.

Does this mean $\|L\|\ne1$?

No, Consider $L: \ell_2 \to \ell_2$. Since $\vert \vert L \vert \vert_{op} = \sup_{\vert \vert x \vert \vert_2  =1} \vert \vert L(x) \vert \vert_2 $

Since $\vert \vert L(x) \vert \vert_2^2 = \sum_{i \geq 2} x_i^2$
and $\vert\vert x \vert\vert_2 = \sum_{i \geq 1}  x_i^2$ it is clear that $ \vert \vert L(x) \vert \vert_2  \leq \vert \vert x \vert \vert_2^2  $

but take $e_2=(0,1,0,0,\ldots)$, $L(e_2) = (1,0,0, \ldots)$ $$\vert \vert L(e_2) \vert \vert_2   = 1 = \vert \vert e_2 \vert \vert_2 $$

so $\vert\vert L \vert\vert_{op} = 1$

remark: $\ell_2 = \{x = (x_1, x_2, \ldots) \mid x_i \in \mathbb{R}, \sum_{i \geq 1}x_i^2 < \infty\}$

It is the adjoint of right shift operator which is a isometry, so it's norm is 1.


Solving Checkers and EXPTIME? In Lance Fortnow's book, The Golden Ticket: P, NP, and the Search for the Impossible, he writes "Computer scientists generally believe that chess, checkers, Othello, and Go are all much harder than any NP problem" (64).

I understand that checkers (as well as the others) are all EXPTIME-Complete, and thus "harder" than an NP-Complete problem, but I don't understand exactly what he means by "harder".

I would assume he means harder to solve, but checkers is a solved game (to force a draw). Am I just not understanding why checkers is categorized as an EXPTIME-Complete game?

In Robson's paper, "N x N Checkers is Exptime Complete", he writes in the Abstract: "The game of Checkers can easily be generalized to be played on an N by N board and the complexity of deciding questions about positions regarded as a function of N. This paper considers mainly the question of whether a particular player can force a win from a given position and also the question of what is the best move in a given position. Each of these problems is shown to be complete in exponential time".

Does that mean that the problem of evaluating a given move in checkers is EXPTIME-Complete, but not the problem of "solving" the game?

The solution of checkers by Schaeffer and his colleagues is for the $8\times 8$ version of the game.  Robson and Fortnow refer to a family of games played by the same rules on $N \times N$ boards with different values of $N$.

An $N \times N$ board can be populated by men and kings in $5^{N^2/2}$ different ways.  This is because there are $N^2/2$ squares where pieces may be placed.  Each of these squares may be in one of five states.)  If we add one bit of state to track which player plays next, we have $2\cdot5^{N^2/2}$ configurations, which for $N=8$ amounts to $46566128730773925781250$.

Many of these configurations are not reachable from the initial configuration, but that's not important.  One can, in principle, draw a graph with a vertex for each configuration, join two vertices with a directed edge if there is a legal move that turns the source configuration into the destination and analyze this graph for a strategy.

Starting from the configurations in which one player has won the game, one can work backwards and identify all other positions from which that player can force a visit of one of its winning positions.  The process can be carried out in time polynomial in the size of the graph.  Once this is done for both players, each configuration is winning for black, winning for white, or drawn.  This is what Robson alludes to with "the proof that the language is in Exptime is simple and is omitted."

While one should not discount the practical difficulty of analyzing the $8 \times 8$ version of the game in a reasonable amount of time, the conclusion that perfect play by both players from the initial position leads to a draw applies to that board size.  In that light, $N \times N$ checkers is not a solved game.  Yes, we have an algorithm that would give us the answer, but it takes impractically long for $N > 8$.

Contrast this situation with the game of nim or with a game I take from Honsberger's More Mathematical Morsels, p. 6.

  From a row of $n \geq 12$ consecutive positive integers, two players, first $A$ and then $B$, take turns crossing out the integer of their choice until there are just two numbers left, $a$, and $b$.  $A$ wins if $a$ and $b$ are relatively prime, and $B$ otherwise.  If $n$ is odd, would you choose to play first or second?  What if $n$ is even?

It turns out that it's easy to analyze the initial configuration to decide which player has a winning strategy and the strategies are simply described.    The amount of work required to analyze an initial position is polynomial in $n$.  (For nim, let $n$ be the total number of counters; it's easy to analyze any position.)

We could, in principle, draw a graph with all possible configurations as vertices and follow the same approach as in $N \times N$ checkers.  For the game from Honsberger's book, a configuration would be a subset of the given set of integers with at least two elements, and a turn value.  However, there are much faster methods, while Robson's proof implies that there is no deterministic polynomial time algorithm to decide whether an arbitrary position of $N \times N$ checkers is winning for a player and we don't know of a fast way to assess the initial position.