Solving a logarithmic equation where the logarhitm is exponentiated I have troubles solving the following logarthitmic equation.

$$ \ 2(\log_x{\sqrt7})^2-\log_x{\sqrt7}-1 =0 $$

The results are supposed to be $ \ x_1 = {\frac{1}{7}}, x_2 = \sqrt7 $

I have tried substitution ( $\log_x{\sqrt7} = x)$, which yielded a quadratic equation with results of $ { x_1 =\frac{1}{2} } $ and $ x_2 = 0 $, none of which works. I would like to ask what the proper way of solving equations like this is?

The key is to substitute $\log_x 7$ with a separate variable (not $x$) and then solve the quadratic

$$ \log_x \sqrt{7} = \frac{1}{2}\log_x 7$$

let $\log_x 7 = a$

Your equation reduces to:

$$2\left(\frac{1}{2}\log_x 7\right)^2 -\frac{1}{2}\log_x 7 - 1 = 0$$

$$ a^2 -a -2 = 0$$

Solve the quadratic and then solve for $x$.

Hint

$\log_x\sqrt{7}$  is a number.

$y$ is a number.

$\gamma$ is a number.

Our equation states that twice a number squared less the number less one is equal to zero.

or
$$2\gamma^2-\gamma-1=0$$
or
$$2y^2-y-1=0$$
or alternatively
$$2(\log_x\sqrt{7})^2-(\log\sqrt{7})-1=0$$


Five members of a wrestling team are weighed and ... An average weight is calculated after each player is weighed. Of the average weight increases by 2 pounds after each player is weighed, how much is the last player than the first?

My answer is 8 pounds. I first mark the five members with letters a to e. Since the difference in weight between a and c is 4 and the difference between c and e is also 4, what I have to do is just add the two 4s up, and get 8 as mu final solution.but I know the answer isn't 8. Could you show me a full algebraic work.

Let's number the players 1 through to 5, and for $1\leq i \leq 5$, let $m_i$ be the weight of player $i$.

We know that $\frac{m_1+m_2}{2} = m_1 + 2$ (the average weight of just one player is just his weight) and hence $m_1+m_2 = 2m_1+4$ so $m_2 = m_1+4$.

Similarly, $\frac{2m_1+4+m_3}{3} = \frac{m_1+m_2+m_3}{3} = \frac{m_1+m_2}{2}+ 2 = m_1 + 4$ so $m_3 = m_1+8$.

Continuing this way and substituting $m_i = m_1 + k$ once we know how much heavier player $i$ is than player 1, we find:

$m_4 = m_1+12$ and $m_5=m_1+16$, so the last player is 16 lbs heavier than the first.

There's a more elegant way to do this, but the above is very explicit and clear.

[Edit:] No, ignore this.  ah11950's correct. I didn't read the question properly.

No, that's it.

The average weight change is the total of the weight changes divided by the number of weight changes.  So the total of the weight changes is the number of weight changes multiplied by the average weight change.

${\bar {w} = \dfrac{\sum\limits_{\forall i} w_i}{n} \iff \sum\limits_{\forall i} w_i = n \bar {w}}$

The only trick here is recognising that there are but four weight changes between the first and fifth player.$$\text{Answer} = 4 \times 2 = 8$$


Integer Square Root Algorithm I'm currently working my way through David M. Bressoud's "Factorization and Primality Testing", and I'm struggling with an exercise (exercise 5.7) that asks the reader to prove that the following algorithm produces the greatest integer less than or equal to the square root of $n$:

\begin{align}
\text{INITALIZE:} \quad &\text{READ} \;n \\
&a \leftarrow n \\
&b \leftarrow \lfloor (n+1)/2 \rfloor \\\\
\text{MYSTERY_LOOP:} \quad &\text{WHILE} \: b < a \; \text{DO} \\
&\quad a \leftarrow b \\
&\quad b \leftarrow \lfloor (a \times a + n) / (2 \times a) \rfloor \\\\
\text{TERMINATE:}\quad &\text{WRITE} \; a
\end{align}

I know from real analysis that the sequence $x_{m+1} = \frac{x_m^2 + n}{2 x_m}$ converges to the square root of $n$ (for any sensible $x_0$), so morally, I can believe that the above algorithm converges to the integer square root of $n$.

How do we formally prove that the above algorithm works (i.e. terminates in a finite number of steps, such that $a^2 \leq n$ and $(a+1)^2 >n$), and moreover, why start the iterations with $b = \lfloor\frac{n+1}{2} \rfloor$? Is $\lfloor\frac{n+1}{2} \rfloor$ just a very crude approximation? What is its significance?

I have tried splitting the floor into a real part minus a part between 0 and 1 (i.e. $\lfloor x \rfloor = x - \{x\}$), but I haven't been able to get very far with this.

In addition, do we have the same quadratic convergence that we get when we compute the (ordinary) square root using Newton's method?

After some rooting around, I have found the answer on another thread, but for the sake of completeness, I will reproduce the argument here (the proof is taken from a textbook of Cohen):

First, note that $a$ and $b$ are always positive. If the algorithm did not terminate, then the condition in the mystery loop ($b <a$) implies that we'd have an infinite, strictly decreasing, sequence of positive integers, which is a contradiction, as there are only finitely many numbers less than $n$. So the algorithm terminates.

Hence, at some point, we must have that $b \geq a$, i.e. $$\left\lfloor \frac{a^2 +n}{2a} \right\rfloor \geq a.$$ Also note that $$\frac{t^2 + n}{2t} \geq \sqrt{n}$$ for all positive $t$ (easily shown), thus we have $$a \geq \left\lfloor\sqrt{n}\right\rfloor.$$ So assume $a$ does not equal $\lfloor\sqrt{n}\rfloor$. So $a \geq \lfloor\sqrt{n}\rfloor +1$ which implies that $a^2 > n$. Putting this together, we get $$0 \leq\left\lfloor\frac{a^2 +n}{2a}\right\rfloor -a = \left\lfloor \frac{n -a^2}{2a} \right\rfloor < 0$$ because $n-a^2 < 0$. This is a contradiction, and thus, $a$ must be equal to the integer square root of $n$.

On the link above, there is also some discussion on the complexity of the above algorithm, which I may add in later.

Using the notation of the $a$ and $b$ is only necessary when programming.

You have correctly identified that the algorithm gives a sequence of values, which I will call (as you did) the $x_m$.

We thus have the relationship $x_{m+1} = \lfloor \frac{x_m^2 + n}{2 x_m} \rfloor$

Starting with $x_0=n$ gives $x_{1} = \lfloor \frac{x_0^2 + n}{2 x_0} \rfloor=\lfloor \frac{n^2 + n}{2 n} \rfloor = \lfloor \frac{n+1}{2} \rfloor$, which answers one of your questions. It also demonstrates that the change from $x_0$ to $x_1$ is always a decrease (unless n=1).

Given two successive terms of the sequence $x_m$ and $x_{m+1}$, there are three possible outcomes:

1) $x_{m+1}=x_m$

2) $x_{m+1}<x_m$

3) $x_{m+1}>x_m$

Let's consider these in turn.

1) $x_{m+1}=x_m$

$\lfloor \frac{x_m^2 + n}{2 x_m} \rfloor=x_m$

The $floor$ function can be dealt with by expressing $\lfloor \frac{x_m^2 + n}{2 x_m} \rfloor=\frac{x_m^2 + n}{2 x_m}-p$ where $0 \le p < 1$

Thus we have $\frac{x_m^2 + n}{2 x_m}-p=x_m$

Rearranging gives $\frac{n-x_m^2}{2 x_m}=p$

Since $0 \le p < 1$, we can say that $0 \le \frac{n-x_m^2}{2 x_m} < 1$

Considering this in two parts, we have first:

$0 \le \frac{n-x_m^2}{2 x_m}$

$x_m$ is positive, so $0 \le {n-x_m^2}$

$x_m^2 \le n$

$x_m \le \sqrt n$

Then:

$\frac{n-x_m^2}{2 x_m} < 1$

${n-x_m^2} < {2 x_m}$

$n<x_m^2 + 2 x_m$

Completing the square,

$n+1<x_m^2 + 2 x_m+1$

$n+1<(x_m + 1)^2$

$\sqrt {n+1}<x_m + 1>$

$\sqrt {n+1}-1<x_m $

Putting the two inequalities back together gives us $\sqrt {n+1}-1<x_m \le \sqrt n$

It can be shown (binomial expansion) that $\sqrt {n+1}-1>\sqrt n -1 $ so we have:

$\sqrt n - 1 <x_m \le \sqrt n$

So if the sequence converges to a value, then this will be the largest integer less than or equal to the square root of $n$.

It could be, however, that the values oscillate without converging. What about those situations?

Moving on:

2) $x_{m+1}<x_m$

Following the same method as before, we have

$\frac{x_m^2 + n}{2 x_m}-p<x_m$

$\frac{n - x_m^2}{2 x_m}<p$

$\frac{n - x_m^2}{2 x_m}<1$

${n - x_m^2}<{2 x_m}$

$n<x_m^2+2 x_m$

$n+1<x_m^2+2 x_m+1$

$n+1<(x_m+1)^2$

$\sqrt {n+1}<x_m+1$

$x_m>\sqrt {n+1}-1$

As before, this gives $x_m>\lfloor \sqrt n \rfloor$

If there are two successive decreases in value, then both $x_m$ and $x_m+1$ will be greater than $x_m>\lfloor \sqrt n \rfloor$, but the values will be moving closer to $x_m>\lfloor \sqrt n \rfloor$, thus having convergence.

3) $x_{m+1}>x_m$

$\frac{x_m^2 + n}{2 x_m}-p>x_m$

$\frac{n - x_m^2}{2 x_m}>p$

We know that $p \ge 0$, so $\frac{n - x_m^2}{2 x_m} \ge 0$

$n - x_m^2 \ge 0$

$x_m^2 \le n$

$x_m \le \sqrt n$

This means that an increase in the values of the sequence only happens if the value in the sequence is less than or equal to the square root. There cannot be an infinite run of increases because eventually the value will become greater than the square root. Therefore there must eventually be a decrease in the values of the sequence.


Drawing a function that isn't $x^3$ with a critical number that is not a local extremum. I have a problem that I'm not sure how to do at all. It says,

  Sketch the graph of a function that is continuous on $[1, 5]$ and has an absolute maximum at $2$, an absolute minimum at $5$, and a critical number at $4$, but no local maximum or minimum there.

I know that $f=x^3$ has a critical number that isn't an extremum, but it doesn't have any absolute extrema either.

I am not sure how to draw a critical point that isn't a local minimum or maximum, and yet still has absolute extrema.

How can I draw a function with absolute extrema, and that has a critical point that isn't a local extremum?

Sketch means to draw a picture describing it.  Not to actually find the expression of one.

Assuming continuity and differentiability then:

An absolute maximum at $2$ means: as $1 < 2 < 5$ it must also be a local maximum so the function "plateaus" at two and the $f'(x)=0$ so it "flattens out" at $x = 2$.

A critical point that is neither a local maximum or minimum means (if it is differentiable) means it "flattens" out at $x=4$ but neither "plateaus" or "valleys".  So it "passes through".

And as 5 is an extreme point of the interval $[1,5]$, an absolute minimum just means the lowest point on the interval. It need not be a critical point.

If you forgive the crappy graphics:

If I have understood your question right then your function should look like this. It should have a Maxima at 2 .minima at 5. It should change its curvature from convex to concave at 4. Notice that these are local Maxima/minima so function can assume values larger/smaller than those at local Maxima/minima


Derivative of $\dfrac{dy^2}{dx}$ I just finish learning the chain rule and am now learning Implicit Differentiation and I am wondering: why is it not possible to take the derivative of $\dfrac{dy^2}{dx}$?  Why do we need to apply the chain rule and find $(\dfrac{dy^2}{dy})\cdot\dfrac{dy}{dx}$?

I understand that $y^2$ is a function and it is equal to $[y(x)]^2$, but basically I am having trouble understanding $\dfrac{dy^2}{dy}$ then $\dfrac{dy}{dx}$.  Why do we find the derivative of ${y^2}$ with respect to $y$, and not $x$?

Note that, if we're working with a function $y=y(x)$, then $y^2$ is just the usual multiplication $y(x)\cdot y(x)=y(x)^2$. This is the same as composing the function
$$
f(x)=x^2
$$
with $y(x)$, which gives $f(y(x))=y(x)^2$.

Now, if we're searching the derivative of $y^2$ with respect to $x$, we must use the Chain Rule and consider $y^2=f(y)$, which gives us
$$
\dfrac{d(y(x)^2)}{dx}=\dfrac{d(f(y(x))}{dx}=\dfrac{df(y)}{dy}\cdot\dfrac{dy}{dx}
$$

And since $\dfrac{df(y)}{dy}=\dfrac{d(y^2)}{dy}=2y$, you get the result. You should notice that since it's an implicit derivative, $\dfrac{dy}{dx}$ will depend on the value of $y(x)$ (that's why it's implicit).

For example, let's suppose that $y(x)$ is given implicitely by the equation $x^2+y(x)^2=1$ and we're trying to find the value of $\dfrac{dy}{dx}$ in a given $x$ point where $y(x)\neq 0$. We derive both sides of the equation, and use the argument given above (before we introduced the example) to get
$$
\dfrac{d}{dx}(x^2+y(x)^2)=0
$$
then
$$
\dfrac{d(x^2)}{dx}+\dfrac{d}{dx}(y(x)^2)=0
$$
$$
2x+2y\dfrac{dy}{dx}=0
$$
This is the part we used the forementioned argument. Now we can get the value we want:
$$
\dfrac{dy}{dx}=\dfrac{-2x}{2y}
$$

I hope it is clear enough!

Given a two fractions, $\frac{a}{b}$ and $\frac{b}{c}$, how do you calculate $\frac{a}{c}$?

Let's relabel $y^2$:

$$\require{cancel}\begin{align}
u&=y^2 \\
\frac{d}{dx}y^2&=\frac{du}{dx}=\cancel{\frac{du}{dy}\frac{dy}{dx}}
\end{align}$$

Take the derivative of $u$ with respect to $y$:
$$\frac{du}{dy}=2y$$


Find the values of $m$, so the integral converges I need to find the values of $m$, so that the following integral converges:
$$\int_{1}^\infty \frac{\ln(1+x)}{x^m}dx$$
I need to choose between two options $m>1$ and $m>0$.
I know that for any $x>e-1,$ $\ln(1+x)>1$, so by the comparison test the integral divergent if $m<1.$
However, I don't know how to prove that it converges for $m>1$.

For $m> 1$, you have when $x$ tends to $+\infty$
$$\frac{\ln(1+x)}{x^m} = o \left( \frac{1}{x^{\frac{m+1}{2}}}\right)$$
Because $\frac{1+m}{2} > 1$, then $\frac{1}{x^{\frac{m+1}{2}}}$ is integrable on $[1, +\infty)$, therefore your integral converges.
For $m \leq 1$,
$$\frac{\ln(1+x)}{x^m} \geq \frac{1}{x^m}$$
which is not integrable, therefore your integral diverges.

We have that for $m>1$
$$\frac{\left(\frac{\ln(1+x)}{x^m}\right)}{\left(\frac{1}{x^{\frac{1+m}2}}\right)}=\frac{\ln(1+x)}{x^\frac{m-1}2}\to 0$$
therefore the integral converges by limit comparison test with $\int_1^\infty \frac{dx}{x^{\frac{1+m}2}}$.
For the case $m\le 1$ we have
$$\frac{\left(\frac{\ln(1+x)}{x^m}\right)}{\left(\frac{1}{x^m}\right)}=\ln(1+x)\to \infty$$
therefore the integral diverges by limit comparison test with $\int_1^\infty \frac{dx}{x^{m}}$.


Maximal tiling without any 3-in-a-rows You are given an arbitrarily large grid, where each square can either be off or on (think Game-of-life type board).

You need to tile such a grid to maximize the number of "on" squares without there being any 3-in-a-row of "on" squares.  A 3-in-a-row can be horizontal, vertical, or diagonal.  3-in-a-row of "off" squares are allowed.

The best tiling I could come up with is

101010101010
101010101010
010101010101
010101010101

which gives a ratio of 1/2.   The best upper bound I have is 6/9 (as you can't fit 7 on any non-wrapping 3x3 square). I believe the optimal solution will be periodic, but if it isn't then that is OK.

Is the tiling above the optimal tiling?  Can this problem be generalized to N-in-a-row?

You can achieve 5/9 density by infinitely tiling this pattern horizontally:

010101
110110
101010

This does not however tile the plane.

I also did an exhaustive search for 3x6, and found that 10/18 was optimal. Any tiling of the plane with a greater density of 10/18 must contain blocks of 3x6 with more than 10 squares on. This is impossible, thus 5/9 is an upper bound on the density.

EDIT: The best bound I've found through exhaustive search is 13/24. No better density is possible.

However @feersum conjectured that for any finite size field $n+1 \over 2n$ is always possible, which if true makes exhaustive search a fruitless effort to prove the bound 1/2 tight.

Optimal or non-optimal, periodic or non-periodic, there's essentially only one way to do this.

Suppose we have an assignment of 0s and 1s to each cell on the infinite grid, not necessarily a tiling (ie. not necessarily periodic), and not necessarily optimal. The trick is to play a kind of Sudoku to show that there is an essentially unique way to do this.

There must be a 1 somewhere in the grid. Now, to the right of this 1, we have three possibilities: either we have 0 0, 0 1, or 1 0. Notice that the case 1 0 is isomorphic to the case 0 0, since in the former case we would have 1 1 0 and in the latter we would have 1 0 0, which are the same thing under the symmetry of flipping the board horizontally and swapping 0s and 1s. So we have only two cases: 1 0 0, and 1 0 1.

Suppose first that there's a 1 0 0 somewhere on the board. We need another case study: either the cell immediately under it is 1, or it is 0. I'll show how you can use the "sudoku" to get a contradiction out of the first assumption:

1 0 0     1 0 0    1 0 0    1 0 0
1      -> 1     -> 1 1   -> 1 1 0   Contradiction!
          0        0        0   0

If the cell immediately under it is 0, then we have the pattern

1 0 0
0

If we try applying the sudoku method to deduce the rest of the grid, we find that we don't run into any contradictions, and indeed we start generating an infinitely tiling pattern:

1 1 0 0
0 0 1 1

You can show that if the grid contains this 2x3 pattern, and there are no three-in-a-rows, then this pattern must tile infinitely in all directions. This completes the case study for when the grid contains a 1 0 0.

Now suppose we have a 1 0 1 somewhere on the grid. Again we have to do a case study: we have either

1 0 1
0

or

1 0 1
1

In either case, calculating outwards as usual, you pretty quickly find the fatal 2x6 pattern mentioned above (possibly rotated 90°), so that's that.


Inequality, what is wrong with $\log(-1) = - \log(-1)$? Can anyone tell me what is wrong with the following line of argument:

$$
 \log(-1) = \log(-1) - \log(1) = - \bigg( \log(1) - \log(-1) \bigg) = - \log \Big( \frac{1}{-1} \Big) = - \log(-1)
$$

Considering the complex logarithm the left-hand-side evaluates to $ i \pi $ and the right-hand-side evaluates to $- i \pi$.

It's because the logarithm, if you extend it to the complex numbers, is only properly defined on a Riemann surface.

What does that mean?  First of all, notice that you're getting $i\pi$ on the left hand side, and $-i\pi$ on the right hand side.  It is interesting to notice that:

$$
e^{i\pi}=-1=e^{-i\pi}
$$

Now the logarithm can be naively defined as the 'inverse of the exponential'.  So why do we say $\log(-1)=i\pi$ and not $\log(-1)=-i\pi$?

Well, first notice that $e^{2i\pi}=1$.  So $e^a=e^{a+2ni\pi}$ for any integer $n$.  This means that the logarithm is only defined up to addition of $2ni\pi$.  There are two ways round this:

Just choose a value for the logarithm at each point.  Remember that the logarithm of a complex number $re^{i\theta}$ is given by $\log_e(r)+i\theta$, where $\theta$ is the argument.  It is common to specify that, for example, $-\pi<\theta\le\pi$.  This is called a branch cut, because we cut along the negative real axis, and end up with one branch of the complex logarithm - another branch is given by $\pi<\theta\le3\pi$.  For the first branch $\theta$ for $-1$ is $\pi$, so we say that $\textrm{Log}(\theta)=i\pi$ (the capital L on $\textrm{Log}$ refers to the principal value of the logarithm - what we get if we restrict ourselves to a particular branch.  Notice that this logarithm coincides with the real logarithm if and only if the argument for real numbers is chosen to be $0$ and not $2n\pi$ for some integer $n\ne 0$.
The branch cut means that the logarithm is not continuous on the negative real axis (indeed, it is usually not defined along the cut at all). Instead of studying the complex logarithm over the complex numbers, we introduce a Riemann surface - a one-dimensional complex manifold that 'joins up' the branch cuts so that the logarithm is now continuous everywhere.  This picture shows you sort of what $\log$ looks like on a Riemann surface:

Unfortunately, in neither of those cases can you assume that usual relationships like $\log(ab)=\log(a)+\log(b)$ still hold.  For example, if we make a branch cut along the negative real axis, we could try to work out $\log((-1+i)\times(-1+i))=\log(-2i)$ by writing it as $2\log(-1+i)$.  The argument $\theta$ for $-1+i$ is $3\pi/2$, but doubling that angle 'crosses' the branch cut, and we end up with $-2i$, which has an argument not of $3\pi$ but of $-\pi$.  Of course, we could place the cut somewhere else and we'd be fine.  In your example, though, you will end up crossing the branch cut wherever you put it.

Why is that?  Your example is basically equivalent to saying that:

$$
\log(-1)=\log\left(\frac{1}{-1}\right)=-\log(-1)
$$

Now $-1$ can be thought of as going round the unit circle by an angle of $\pi$.  The reciprocal of a complex number always has negative of the argument of the original number, so $\dfrac{1}{-1}$ can be thought of as going round the unit circle by an angle of $-\pi$; i.e., you reach the same point travelling in the opposite direction.  Between them, the two paths cover the entire unit circle.  Since you've got to place a branch cut somewhere on the circle, you're going to end up crossing it at some point, so you've got to get an answer that makes no sense.

Update: I can't remember where I first read this, but it's definitely true.  It's easy to look at something like this and conclude that it's a slightly annoying area of mathematics that you have to deal with even though you'd rather it worked a different way.  In fact, the opposite is true: this seemingly annoying property of the complex logarithm in fact opens up many beautiful areas of mathematics, including the study of Riemann surfaces.

I think $\log(1)=0$ doesn't hold in the complex,so "$\log(-1)=log(-1)-log(1)$"is wrong.


The boundary of a manifold has empty boundary It is a known fact, for example well explained here, that the boundary of an $n$ manifold is an $n-1$ manifold, and the proof shows essentialy that properly restricted charts which parametrize $M$ parametrize also $\partial M$, with open set of a lower dimension, $n-1$.
My question is : I have to show that $\partial M$ has no boundary, or is implicit somehow having shown that $\partial M$ is parametrizable with open sets of $\mathbb{R}^{n-1} ?$ If so, why ?
Because for my definition of boundary I don't see how this should be implicit.
My definition is $p \in \partial M \iff p \in \phi(U \cap \left\lbrace x_n = 0\right\rbrace)$ i.e is the image of $\partial \mathbb{H}^n \cap U$, (where $U$ is the open set of $\mathbb{R}^n$ such that the intersection with closed half space $U \cap \mathbb{H}^n$ is a diffeomorphism with an open set of the manifold $V$).
How should I prove it ? It could be useful using this fact ?
I'd like to understand the answer given in this previous question, which seems the way I should prove, it but I don't understand how the fact follows easily from the definitions.
Any help would be appreciated.

By definition, you $\partial M$ is endowed with charts that "look like" an open subset $U\cap \Bbb R^{n-1}$.


Linear independence and the Wronskian Suppose I have two linearly independent solution vectors \begin{bmatrix}x_1,_1(t)\\x_1,_2(t)\end{bmatrix} and \begin{bmatrix}x_2,_1(t)\\x_2,_2(t)\end{bmatrix}

If I take the Wronskian of these 2 solution vectors, it comes out to a nonzero number since they are stated to be Linearly Independent. My question is, if you take the Wronskian of the same solution vectors but their derivative:

\begin{bmatrix}x'_{1,1(t)}&x'_{2,1(t)}\\x''_{1,1(t)}&x''_{2,1(t)}\end{bmatrix}

Would it still be linearly independent? (Would the Wronskian still be a nonzero number?

You've got a couple things going on at once here. First, you are asking about the Wronskian of two vectors of functions:
$$\mathbf{x}_1(t) = \left[\begin{array}{c} x_{1,1}(t) \\ x_{1,2}(t) \end{array}\right] \quad\text{and}\quad \mathbf{x}_2(t) = \left[\begin{array}{c} x_{2,1}(t) \\ x_{2,2}(t) \end{array}\right]$$
which is defined as:
$$W(\mathbf{x}_1,\mathbf{x}_2)(t) = \left|\begin{array}{cc} x_{1,1}(t) & x_{1,2}(t) \\ x_{1,2}(t) & x_{2,2}(t) \end{array}\right|.$$
Next you are asking about the Wronskian of two differentiable functions $x_{1,1}(t)$ and $x_{2,1}(t)$, which is defined as:
$$W(x_{1,1},x_{2,1})(t) = \left|\begin{array}{cc} x_{1,1}(t) & x_{2,1}(t) \\ x_{1,1}'(t) & x_{2,1}'(t) \end{array}\right|.$$
So be careful there. To answer your question (for functions at least), take:
$$f(t) = t\quad\text{and}\quad g(t) = -1.$$
Then,
$$W(f,g) = \left|\begin{array}{cc} t & -1 \\ 1 & 0 \end{array}\right| = 1$$
but
$$W(f',g') = \left|\begin{array}{cc} 1 & 0 \\ 0 & 0 \end{array}\right| =0.$$
Also, this link does a great job at explaining the Wronskian.

Any two linearly independent constant vectors have dependent (identically zero) derivatives. Thus, the answer to your question is "not necessarily".


I am having difficulty with the algebra related to a coordinate geometry question. A line is drawn through the point $A=(1,2)$ to cut the line $2y = 3x-5$ in $P$ and the line $x+y = 12$ in $Q$.  If $AQ$ = $2AP$, find the coordinates of $P$ and $Q$.

Case 1:

Since $Q(a,12-a)$ is on line $x+y=12$ then the midpoint $P({a+1\over 2},{14-a\over 2})$ of $AQ$ is on line $2y=3x-5$. So we have:

$$2{14-a\over 2} =3{a+1\over 2} -5$$

So we have $a= 7$ and thus $Q(7,5)$ and $P(4,{7\over 2})$.

Case 2:

The midpoint $P'({a+1\over 2},{14-a\over 2})$ of $AQ$ is not on line $2y=3x-5$.
Since $A$ is the midpoint of $PP'$ we have $P({3-a\over 2},{a-6\over 2})$. Since $P$ is on $2y=3x-5$ we have:

$$2{a-6\over 2} =3{3-a\over 2}-5$$

so $a= {5\over 4}$...

Hint -
As AQ=2AP hence AP=PQ. Therefore use midpoint formula
Now consider P($x_1 , y_1$) and Q($x_2 ,y_2$).
Now slope AP= slope PQ = slope AQ . Now  P($x_1 , y_1$) satisfy equation 1 and and Q of another one. I hope you can solve forward.
Remember it's just a hint. Next time write what y you have tried


What is $\lim_{x \to 0} \frac{\ln(x^{{10}^8})}{x^2}$? What does this evalute to?

$$\lim_{x \to 0} \frac{\ln(x^{{10}^8})}{x^2}$$

I know basic facts relating to limits of $\ln$ and $x$, and WolframAlpha says the answer is $-\inf$. I can convince myself that this is true, but how can one show it?

Thanks.

You have $\lim_{x \to 0^{+}} \log x = -\infty$ and $\lim_{x\to 0^{+}} \frac{1}{x^2} = \infty$ so $\lim_{x \to 0^{+}} \frac{\log x}{x^2} = -\infty$.

What this is really saying is that $\log x$ is unbounded below as $x\to 0^{+}$ and $1/x^2$ is unbounded above as $x\to 0^{+}$ so their product is also unbounded below as $x\to 0^{+}$.

Edit: as per Jonas's comment, you can replace all instances of $\lim_{x\to 0^{+}} \log x$ with $\lim_{x\to 0} \log |x|$ since $\log (x^{10^8}) = 10^8 \log  |x|$.

Note $\mathrm{ln}(x^{10^8}) = 10^8\mathrm{ln}(x)$ so ignore the $10^8$ for the moment and just use L'Hospital's rule on $\frac{\mathrm{ln}(x)}{x^2}$.

Edit: Oops, I misread the limit as going to $\infty$.  In fact the limit is going to $0$, in which case both $-\mathrm{ln}(x^{10^8})$ and $\frac{1}{x^2}$ go to $\infty$, so their product goes to $\infty$, so $\frac{\mathrm{ln}(x^{10^8})}{x^2}$ goes to $-\infty$.


Area between two polar curves $r = 2 \sin\theta$ and $r =2\cos\theta$ I am trying to find the area between the polar curves $r = 2 \sin θ$ and $r = 2 \cos θ$.

I set up the area equation as follows:

$$\frac12\int_0^{\pi/4}((2\sinθ)^2-(2\cosθ)^2)\,d\theta$$

I could not get the correct answer with this, which is $\frac\pi2-1$.

Any help with this problem would be appreciated :D

A picture might help: $r=2\sin t$ is in blue and $r=2\cos t$ is in red for $0\le t\le 2\pi$.

The intersection point is found by setting $2\sin t=2\cos t$ and solving; it occurs at $t=\pi/4$, so you want
$$
\underbrace{{1\over 2}\int_0^{\pi/4} (2\sin t)^2\,dt}_\text{area of blue region}
+
\underbrace{{1\over 2}\int_{\pi/4}^{\pi/2} (2\cos t)^2\,dt}_\text{area of red region}={\pi\over 2}-1.
$$

this is just a hint and you have to do it by yourself so see below.....
by using 12th method
                so
we know by polar co-ordinates
   $x= r \sin(x)$  and $y=r \cos(x)$
    here   in question

$r=2\sin(x)$

multiply both side $r$

$r^2=2r\sin(x)$

$r^2 = 2x$   since $x=r \sin(x)$

since a point circle equation is $x^2+y^2=r^2$

therefore
$x^2+y^2=2x $ implies that   $y=\sqrt (2x-x^2)$
 similarly
for $r=2\cos(x)$,
 $x^2+y^2=2y$  implies that   $y= \sqrt (2y-x^2)$
and integrate it by drawing the circle

get the answer.
best of luck


Avoiding collinearity in logistic regression I have the following problem: I'm performing a multivariate logistic regression on several variables each of which has a nominal scale. I want to avoid multicollinearity in my regression. If the variables were continuous I could compute the variance inflation factor (VIF) and look for variables  with a high VIF. If the variables were ordinally scaled I could compute Spearmon's rank correlation coefficients for several pairs of variables and compare the computed value with a certain threshold. But what do I do if the variables are just nominally scaled? One idea would be to perform a pairwise chi-square test for independece, but the different variables don't all have the same codomains. So that would be another problem. Is there a possibility for solving this problem?

Thanks in advance!

Strictly speaking, multicollinearity is a concept which is unique to real-valued covariates. It's also a specialized indicator of dependence between covariates. You have the right idea in that it's important to avoid covariates that are too "mutually dependent" - even if they have categorical domains. In fact, you can actually use the idea of multicollinearity to motivate a statistic to measure such dependence among categorical covariates. I'll try to do this through an example.

Suppose you have two covariates, $X_1$ and $X_2$, which are both continuous. The VIF is equal to $1/(1-R^2)$, where $R^2$ is the squared-correlation between $X_1$ and $X_2$. Thus, multicollinearity simply depends how well $X_1$ predicts $X_2$ as assessed by $R^2$.

Now, suppose $X_1$ and $X_2$ are categorical. You can still use $X_1$ to predict $X_2$ (e.g. through a logistic regression). But you wouldn't assess the quality of the prediction using the usual $R^2$ in this case. There are many different pseudo-$R^2$ formulas you could consider. I think one useful approach would be to consider one based on the entropy reduction. For instance, the marginal entropy, $H(X_2)$ of $X_2$ can easily be estimated. Additionally, the conditional entropy, $H(X_2|X_1)$ can be estimated nearly as easily. You can then develop a pseudo-$R^2$ with the formula,
$$R^2_{pseudo} = 1 - \dfrac {\hat{H}(X_2|X_1)} {\hat{H}(X_1)}$$
Large values of $R^2_{pseudo}$ would then imply increased dependence among the covariates and small values would indicate otherwise.

The vif function in the car package for R implements generalized variance inflation factors. The documentation references [1]. This function can be used in conjunction with a GLM---I've just tried it on a logistic regression with multiple categorical independent variables and spits out numbers.

> library(car)
> m = glm(am~cyl+vs+carb+hp, data=mtcars, family=binomial)
Warning message:
glm.fit: fitted probabilities numerically 0 or 1 occurred
> car::vif(m)
         cyl           vs         carb           hp
2.017467e+07 2.017465e+07 5.926166e+00 1.639876e+01

[1] Fox, J. and Monette, G. (1992) Generalized collinearity diagnostics. JASA, 87, 178–183.


Markov chain - Regular transition matrix I have to prove that this transition matrix is regular but how can I prove it without having to multiply it n times?
I know when you raise it to the 2nd power you have to check if the zeros in the first matrix are at the same entries as in the second one but anyone has a more in depth explanation?

A stochastic matrix is regular if it's irreducible and has at least one non-zero entry on its main diagonal.  It's easy to show that that your matrix is irreducible, since every state communicates with state $1$, and state $\ i\ $ communicates with state $\ i+1\ $ for $\ i=1,2,3,4\ $, and the first entry on its main diagonal is non-zero.  Therefore it's regular.


Roots of the equation $|x|^{n-2}x^2 +\lambda = 0$, $\lambda\in(0,\infty)$ I want to find all the roots of the equation $$|x|^{n-2}x^2  +\lambda = 0,$$ where $\lambda\in (0,\infty)$.

I'd like to know how I can find them.

Note When this answer was written, the OP's question was to solve the equation where $\lambda\equiv 1$. The OP since changed the question after this answer was given.

Clearly $x\neq 0$ for otherwise $1=0$ which is absurd. Hence, let $x=re^{i\theta}$ where $r>0$, then

$$|re^{i\theta}|^{n-2}r^2e^{2i\theta}+1=0,$$
giving
$$|r|^{n-2}|e^{i\theta}|^{n-2}r^2e^{2i\theta}+1=0.$$
Since $r>0$ and $|e^{i\theta}|=1$, then
$$r^ne^{2i\theta}+1=0.$$
Hence,
$$r^n\cos(2\theta)+ir^n\sin(2\theta)+1=0.$$
We obtain the system of equations,
$$\left\{\begin{array}{rll}r^n\cos(2\theta)+1&=&0\\\sin(2\theta)&=&0\end{array}\right.$$

From the sine equation we see that $\theta=\pi k$ or $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$, in which case $\cos(2\theta)$ equals $1$ and $-1$ respectively.

Hence, when $\cos(2\theta)=1$, then $r^n=-1$, which is absurd since $r>0$ and $n\in\mathbb{Z}$. When $\cos(2\theta)=-1$, then $r^n=1$ which implies $r=1$.

So, the only solutions are $r=1$ and $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$.

Therefore, $$x=re^{i\theta}=1\cdot e^{i\frac{\pi}{2}(2k+1)}=\pm i,$$
as you quite rightly pointed out.

UPDATE

The only solution for $\lambda=0$ is $x=0$. Suppose $\lambda>0$, then we obtain the slightly different system of equations,
$$\left\{\begin{array}{rll}r^n\cos(2\theta)+\lambda&=&0\\\sin(2\theta)&=&0\end{array}\right.$$

From the sine equation we see that $\theta=\pi k$ or $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$, in which case $\cos(2\theta)$ equals $1$ and $-1$ respectively.

Hence, when $\cos(2\theta)=1$, then $r^n=-\lambda$, which is absurd since $r,\lambda>0$ and $n\in\mathbb{Z}$. When $\cos(2\theta)=-1$, then $r^n=\lambda$ which implies $r=\lambda^{1/n}$.

So, the only solutions are $r=\lambda^{1/n}$ and $\theta=\frac{\pi}{2}(2k+1)$ where $k\in\mathbb{Z}$.

Therefore, $$x=re^{i\theta}=\lambda^{1/n}\cdot e^{i\frac{\pi}{2}(2k+1)}=\pm \lambda^{1/n}i.$$

Note the following $$x^2 = |x|^2 $$ Therefore the equation becomes $|x|^n=-1$.

$-1 = e^{(2k-1)i\pi}\hspace{1cm}\text{ for k in {1,2..n}}$

So the roots are $$|x| = e^{\frac{(2k-1)i\pi}{n}}\text{for k in {1,2,..n}}$$ or $$x = \pm e^{\frac{(2k-1)i\pi}{n}}\text{for k in {1,2,..n}}$$


Extrema of this sequence I have to find the extrema of:
$$a_n=(-1)^{n+1}\left(2-\frac{n+2}{2^n}\right)$$
I calculated $a_1= \frac{1}{2}$ and $\displaystyle \lim_{n} a_n=(-2,2)$.
Then
$$\left|\left(2-\frac{n+2}{2^n}\right)\right|<2$$

thus I think that $\sup=2$ and $\inf=-2$ but I am not sure how to demonstrate it.

My attempt:

I separate $a_n$ into $a_{2k}$ and $a_{2k-1}$, then:

$$a_{2k}<2 \Rightarrow 1\left(2-\frac{2k+2}{2^{2k}}\right)<2, \text{ true for } k>-1$$

$$a_{2k-1}>-2 \Rightarrow -1\left(2-\frac{2k+1}{2^{2k-1}}\right)>-2, \text{ true for } k>-\frac1{2}$$

So I know that $a_n$ is bounded.

I tried to demonstrate that $\sup=2$ and $\inf=-2$ by doing the following:

$$a_{2k}>2-\varepsilon \Rightarrow 1\left(2-\frac{2k+2}{2^{2k}}\right)>2-\varepsilon$$

$$a_{2k-1}<-2+\varepsilon \Rightarrow -1\left(2-\frac{2k+1}{2^{2k-1}}\right)<-2+\varepsilon,$$
but I can't solve those inequalities. Is this one the right approach?

EDIT

Another attempt: (Thanks to G Cab's answer)

$$-b_n<a_n<b_n$$

where $b_n=\left(2-\frac{n+2}{2^n}\right)$.

$-b_n$ is decreasing and the limit is $-2$, and as the sequence is decreasing, $-2=\inf(-b_n)$.

$b_n$ is increasing and the limit is $2$, and as the sequence is increasing, $2=\sup(b_n)$. Thus

$$-2<-b_n<a_n<b_n<2$$

and $\sup(a_n)=2$ and $\inf(a_n)=-2$

Note that
$$
\eqalign{
  & f(x) = 2 - {{x + 2} \over {2^{\,x} }}\quad  \Rightarrow \quad y(0) = 0  \cr
  & f'(x) = {{\ln 2\left( {x + 2} \right) - 1} \over {2^{\,x} }}  \cr
  & 0 < f'(x) = \quad  \Rightarrow \quad {1 \over {\ln 2}} - 2 \approx  - 0.557 < x \cr}
$$
so $f(n)$ is increasing for $0 \le n$.

Thus the alternating sequence $(-1)^{n+1}f(n)$ is clearly bounded to stay between $\pm 2$.

I am not a theorist, but I think that a formal demonstration should develop along these lines
(maybe somebody more expert can help and put it in right terms)

Given that
$$
\left\{ \matrix{
  a_{\,n}  \in Q = \left( { - 1} \right)^{\,n + 1} f(n) \hfill \cr
  \left\{ {a_{\,n} } \right\} = \left\{ {a_{\,2n} } \right\} \cup \left\{ {a_{\,2n + 1} } \right\}\quad \left| {\;\left\{ {a_{\,2n} } \right\} \cap \left\{ {a_{\,2n + 1} } \right\} = \emptyset } \right. \hfill \cr}  \right.
$$
and that
$$
\eqalign{
  & \mathop {\lim }\limits_{n\; \to \;\infty } a_{\,2n}  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - 1} \right)^{\,2n + 1} f(2n) =  - \mathop {\lim }\limits_{x\; \to \;\infty } f(x) =  - 2  \cr
  & \mathop {\lim }\limits_{n\; \to \;\infty } a_{\,2n + 1}  = \mathop {\lim }\limits_{n\; \to \;\infty } \left( { - 1} \right)^{\,2n + 2} f(2n + 1) = \mathop {\lim }\limits_{x\; \to \;\infty } f(x) = 2 \cr}
$$
and having demonstrated that $f(x)$ is increasing
$$
f(n - 1) < f(n) < f(n + 1)
$$
that means
$$
\eqalign{
  &  - 2 < a_{\,2n} \, \le a_{\,0}  =  0  \cr
  & 1/2 = a_{\,1}  \le a_{\,2n + 1} \, < 2 \cr}
$$
and therefore
$$
\eqalign{
  & \inf \left\{ {a_{\,n} } \right\} = \min \left( {\inf \left\{ {a_{\,2n} } \right\},\inf \left\{ {a_{\,2n + 1} } \right\}\;} \right) =  - 2  \cr
  & \sup \left\{ {a_{\,n} } \right\} = \max \left( {\sup \left\{ {a_{\,2n} } \right\},\sup \left\{ {a_{\,2n + 1} } \right\}\;} \right) = 2 \cr}
$$

I think, you are right because $$\lim_{n\rightarrow+\infty}a_{2n}=-2$$ and

$$\lim_{n\rightarrow+\infty}a_{2n-1}=2$$ and use your work.


Using integral definition to solve this integral I'm trying to solve this question using the definition of integral:

  $$\int^5_2 (4-2x)dx$$

Definition of integral:

We define first the inferior and superior sum:

Let $f:[a,b]\to \mathbb R$ be a bounded function and $P=\{t_0,\ldots,t_n\}$ a partition of [a,b], so we define:

$$s(f;P)=\sum_{i=1}^nm_i(t_i-t_{i-1})$$

$$S(f;P)=\sum_{i=1}^nM_i(t_i-t_{i-1})$$

Where $m_i=\inf\{f(x);x\in [t_i,t_{i-1}]\}$ and $M_i=\sup\{f(x);x\in [t_i,t_{i-1}]\}$ and ().

No we define the inferior and superior integral:

Inferior integral: $\underline\int^b_a=\sup_P s(f;P)$

Superior integral: $\bar\int^b_a=\inf_P s(f;P)$

At last, we define the integral being the superior or inferior integral when they are the same.

My attempt

I know:

$s(f,P)=\sum_{i=1}^n(4-2t_i)(t_i-t_{i-1})$ (because the function is decreasing)

But I don't know how to take the supreme of this sum.

Thanks.

The contributions from fellow users Mhenni Benghorbal and oks pave the way.

Using equidistant spacings one can perform the summation like this:
$$
\begin{align}
s(f,P') &= \sum_{i=1}^n(4-2t_i)(t_i-t_{i-1}) \\
&= \left(\sum_{i=1}^n 4 - 2 \left(a + \frac{b-a}{n} i \right) \right)\frac{b-a}{n} \\
&= \left(4n - 2 \left(a n + (b-a) \frac{n+1}{2} \right) \right) \frac{b-a}{n} \\
&= \left(4 - 2a - (b-a)\left(1+ \frac{1}{n}\right) \right) (b-a) \\
&= \left(4 - 2a - (b-a)\right) (b-a) - \frac{(b-a)^2}{n} \\
\Rightarrow \sup s(f, P') &= \left(4 - 2a - (b-a) \right) (b-a) \\
&=4(b-a) -(b+a)(b-a) \\
&=4(b-a) -(b^2 - a^2)
\end{align}
$$
and
$$
\begin{align}
S(f,P') &= \sum_{i=1}^n(4-2t_{i-1})(t_i-t_{i-1}) \\
&= \left(\sum_{i=1}^n 4 - 2 \left(a + \frac{b-a}{n} (i-1) \right) \right)\frac{b-a}{n} \\
&= \left(4n - 2 \left(a n + (b-a) \frac{n-1}{2} \right) \right) \frac{b-a}{n} \\
&= \left(4 - 2a - (b-a)\left(1-\frac{1}{n}\right) \right) (b-a) \\
&= \left(4 - 2a - (b-a)\right) (b-a) + \frac{(b-a)^2}{n} \\
\Rightarrow \inf S(f,P') &= \left(4 - 2a - (b-a) \right) (b-a) \\
&=4(b-a) -(b^2 - a^2)
\end{align}
$$

Hint: if you are integrating over an interval $[a,b]$ then

  $$ \Delta t_i = t_i-t_{i-1} = \frac{b-a}{n}.  $$

Also, your function is a decreasing function then you should know what $m_i$ and $M_i$ are.


Inverse function of an integral function I am trying to find the inverse function of the following function , or prove that it does not have one , but I can't do either of those things.

$$ y(τ)= \int_{τ-1}^{τ+1} \cos(\frac{πt}{8})\,x(t) \,dt $$ where $x(t)$ is a function with the same domain as $y(t)$. Can someone help with finding its inverse function or proving it doesn't have one . Proving that it has one, can also help.

After evaluating the integral, we can see $y(t)$ is constant $\forall \ t$. Therefore, it has no inverse.


multiplicity of the root of $g(x)=f(x)/f'(x)$, given a root of $f(x)$ and $f'(x)$. Here is a quote from my textbook:

"If $f(x)$ has a root at $x=\alpha$ with multiplicity $m>1$, then $f'(x)$ has a root at $x=\alpha$ with multiplicity $m-1$. Then, the function
\begin{align*}
g(x)=f(x)/f'(x)
\end{align*}
has a root at $x=\alpha$ with multiplicity 1 (a simple root)."

But how is $\alpha$ a root for $g$, since $g(\alpha)=0/0$? I'm quite confused, any help appreciated.

Write $f(x)=(x-a)^m h(x)$, with $h(a)\ne0$. Then
$$
g(x)
= \frac{f(x)}{f'(x)}
= \frac{(x-a)^m h(x)}{m(x-a)^{m-1}h(x)+(x-a)^m h'(x)}
= \frac{(x-a) h(x)}{mh(x)+(x-a)h'(x)}
= (x-a)j(x)
$$
where
$$
j(x)=\frac{h(x)}{mh(x)+(x-a)h'(x)}
$$
Since $j(a)=\dfrac{h(a)}{m h(a)}=\dfrac{1}{m}\ne0$, we get that $a$ is a simple root of $g$.


An inequality in Lemma 4.3 of "PRIMES is in P" by AKS I'm having trouble verifying the following inequality in Lemma 4.3 in the paper PRIMES is in P.

\begin{align}
&n^{\lfloor\log{\lceil\log^5n\rceil}\rfloor+\frac{1}{2}(\log^4(n)+\log^2(n))} < n^{\log^4(n)}\\
&\text{where }n>2\text{ and }n \in\mathbb{N}
\end{align}

The same inequality is dealt with differently in the older Annals of Mathematics version, but it contains a mistake as pointed out here, which is corrected in the newer version that is cited above.

I have used a graphing software to verify the inequality but am unsure as to how to proceed.

Thanking you in advance!

Edit: the base of the logarithm is $2$.

Since $f(x) = n^x$ is a strictly increasing function for positive $x$, or you can see this by taking logarithms of both sides, the equality you want to prove is equivalent to

$$\lfloor\log{\lceil\log^5n\rceil}\rfloor+\frac{1}{2}(\log^4(n)+\log^2(n)) \lt \log^4(n) \tag{1}\label{eq1A}$$

Note that, as stated in the linked paper in section $3$, "Notation and Preliminaries",

  We use log for base $2$ logarithm, and ln for natural logarithm.

One way to show \eqref{eq1A} holds is to break the inequality into $2$ parts. First, consider

$$\frac{1}{2}\left(\log^2(n)\right) \lt \frac{1}{6}\left(\log^4(n)\right) \iff 3 \lt \log^2(n) \tag{2}\label{eq2A}$$

This is satisfied for $n = 4$, and all larger values since $\log(n)$ is a strictly increasing function for positive $n$.

Next, have

$$y = \log^4(n) \tag{3}\label{eq3A}$$

so $y = 16$ for $n = 4$, and

$$\begin{equation}\begin{aligned}
\lfloor\log{\lceil\log^5n\rceil}\rfloor & \lt \frac{1}{3}\left(\log^4(n)\right) \\
\lfloor\log{\lceil y^{\frac{5}{4}}\rceil}\rfloor & \lt \frac{y}{3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$

Consider

$$y^{\frac{5}{4}} + 1 \lt y^{\frac{4}{3}} \tag{5}\label{eq5A}$$

This is true for $y = 16$ and, since $f_1(y) = y^{\frac{4}{3}}$ is a faster growing function than $f_2(y) = y^{\frac{5}{4}}$, \eqref{eq5A} is true for all $y \ge 16$. Next, consider

$$\begin{equation}\begin{aligned}
\lfloor\log{\lceil y^{\frac{5}{4}}\rceil}\rfloor & \lt \lfloor\log(y^{\frac{5}{4}} + 1)\rfloor \\
& \lt \lfloor\log(y^{\frac{4}{3}})\rfloor \\
& \le \frac{4}{3}(\log(y))
\end{aligned}\end{equation}\tag{6}\label{eq6A}$$

As such, you can prove that \eqref{eq5A} is to have the RHS of \eqref{eq6A} be less than or equal to the RHS of \eqref{eq5A}, i.e.,

$$\frac{4}{3}(\log(y)) \leq \frac{y}{3} \iff 4\log(y) \leq y \tag{7}\label{eq7A}$$

For $y = 16$, the $2$ sides of \eqref{eq7A} are equal. Since, for $y \ge 16$, $f_3(y) = y - 4\log(y) \implies f'_3(y) = 1 - \frac{1}{\ln(2)y} \gt 0$, this shows \eqref{eq4A} is true for all $y \ge 16$.

Finally, using \eqref{eq2A} and \eqref{eq4A} in the LHS of \eqref{eq1A} shows it's less than the RHS, so \eqref{eq1A} will be true for all $n \ge 4$. Since you're considering $n \gt 2$ and $n \in \mathbb{N}$, this just leaves $n = 3$ to check, either through your graph or just doing it manually.


Is there an algorithm which solves the puzzle game Mummy Mystery? I had the puzzle board game Mummy Mystery from Maze Ways (a version of the game GoGetter) as a kid, and just out of curiosity I tried my hand at making a computer program which will generate solutions to any of the individual puzzles included with it. For those not familiar with the game: it consists of nine tiles, each depicting the segments of a pathway, and a 3x3 grid with various characters drawn on the outside edges. The goal of each puzzle is to arrange the tiles on the grid, in any order and orientation, so that certain characters are connected via complete pathways (and sometimes the goal is to avoid connecting them). In the classical version of the game, there are no 'dead ends' allowed; the pathway can't simply stop in the middle of the grid and not be continued on the adjacent tile. Some versions will contain other unique features, but this is the basic premise.

Does anyone know of an algorithm which is able to help solve this type of puzzle, or a similar one, in a time-efficient way? I think I'm close to finding a way of generalizing the puzzle which might allow it to be solved, but it involves assigning prime numbers to each pair of nodes and dealing with massive numbers which probably aren't compatible with my computer. I have a very basic understanding of group theory, but I'm unsure how it could be used to help in this situation.

There's a very simple polynomial time algorithm: you try placing the tiles in all possible positions and orientations, and see whether each one achieves the goal. When you find one that does, you stop. This is an O(1) algorithm. The constant (which is larger than $9! \cdot 9^4$, which is simply the number of possible position-orientation layouts, and doesn't assign any time for checking what's connected to what) is large, but that's life.


$∅\in\{∅,\{∅\}\}$ is false but $∅\in\{∅\}$ is true... why? I could be wrong, but I learned that $∅\in\{∅,\{∅\}\}$ is false but $∅\in\{∅\}$ is true... why is that? I would think that both of these would have to be true. If $∅\in\{∅\}$ is true wouldn't $\{∅\}$ be an element in $\{∅,\{∅\}\}$?

Both of these statements are correct as you have suggested.
In both of your sets the empty set appears as an element so there should not be any confusion.

since there is one comma in the set it indicates that there are two different objects in the set, hence empty set symbol as one object, and the brace that  contains the empty set symbol as another object,therefore, both are elements of the set you constructed and hence true correct, and in similar argument for the second case you said true also I agree true because the set contains the empty set symbol as one object only in it. After you enter inside set only find definite objects separated by comma only to consider which definite objects are the members of the set whatever that definite object is.


Denesting Phi, Denesting Cube Roots I have been looking into denesting square roots but I have found that $\sqrt[3]{2+\sqrt{5}}$ equals $(1+\sqrt{5})/2$. The same is true for  $\sqrt[3]{2-\sqrt{5}}$ and $(1-\sqrt{5})/2$. I cannot figure how this is true. I proved this by setting both equal to x and forming polynomial equations. $y= x^2-x-1$ and $y=x^6-4x^3-1$. I do not know how to solve a degree 6 polynomial by hand like that. My objective is to denest the original  $\sqrt[3]{2+\sqrt{5}}$ without knowledge of the simplified phi.

Hint : As a general rule, when dealing with nested radicals of the form $\sqrt[n]{A+B\sqrt[m]C}$ , you write $A+B\sqrt[m]C=(a+b\sqrt[m]C)^n$, and then employ Newton's binomial theorem. In our case, we have

$$\left(a+b\sqrt5\ \right)^3=a^3+3a^2(b\sqrt5)+3a(b\sqrt5)^2+(b\sqrt5)^3=\underbrace{(a^3+15ab^2)}_2+\underbrace{(3a^2b+5b^3)}_1\sqrt5$$ $$\iff a^3+15ab^2=2(3a^2b+5b^3)\iff a^3+15ab^2-6a^2b-10b^3=0\quad|:b^3\iff$$ $$\iff\left(\frac ab\right)^3-6\left(\frac ab\right)^2+15\left(\frac ab\right)-10=0\iff x^3-6x^2+15x-10=0\iff x=1$$ $$\iff\frac ab=1\iff a=b\iff a^3+15a^3=2\iff16a^3=2\iff a=\sqrt[3]\frac18=\frac12$$ or $3b^3+5b^3=1\iff b=\sqrt[3]\frac18=\frac12$ . Similarly for when the second term is $-1$.

It is so easy to solve x6−4x3−1=0
just let x3=t and refine the equation as t2-4t-1=0
after you find the t, you can bring back the x3
for example if t=5, then x3=5 and x=3√5


If $v=dx/dt$ what is $v^2$ equal to? Just working on mechanics problems right now and I realized that I never actually dealt with a problem like this before.

Usually when I'm solving a differential equation, I would often be able to convert variable $v$ to $dx/dt$. However, this question requires me to convert $v$ into a form related to $dx/dt$.

Well... like that? $$(v = \frac{\mathrm{d}S }{\mathrm{d} t}) \xrightarrow{f: x \mapsto x^2} v^2 =  \frac{\mathrm{d}S }{\mathrm{d} t} \frac{\mathrm{d}S }{\mathrm{d} t} \Leftrightarrow v^2 =  \frac{\mathrm{d}S }{\mathrm{d} t^2} \mathrm{d}S $$

which give us $$v^2 = a * \mathrm{d}S$$

I'm probably wrong, but I though I could give it a try.


Proving $f(f^{-1}(x))=x$ While reading about inverse functions I found a statement showing "$f\{f^{-1}(x)\}=x$". I know it is correct but I tried to prove it by taking an simple example.

Suppose there is a function $y=f(x)$ given as, $f:$ $\ce{A=B}$. Here $A$ and $B$ represents domain and range respectively and are sets given as $A=\{1\}$ & $B=\{2\}$, then $f(x)$ can be also written as $f(x)=\{(1,2)\}$. So for $x=1$ (domain), $y=2$ (range). Where $1$ belongs to $A$ and $2$ belongs to $B$.

Also $f^{-1}(x)=\{(2,1)\}$.
Now $f\{f^{-1}(x)\}=\{(2,2)\}$, i.e. identity function of range $y$, i.e. set $B$. As in identity function range $=$ domain. Therefore $y=x$. Hence $f\{f^{-1}(x)\}=x$ Am I correct in explaining this? As I think my method of explanation is wrong. Please tell me the right one & please point out my mistakes.

Let $f:X \to Y$ and $A \subseteq Y$. The preimage of a set is defined as follows:

$$f^{-1}(A)=\{x \in X \mid f(x) \in A\}.$$

This means that the preimage of $\{x\}$ (i. e. $f^{-1}(\{x\})$ are exactly the elements so that $f$ applied them lands you in $\{x\}$, as in $f(f^{-1}(\{x\}))=x.$

Consider $f:A\longrightarrow B$ such that given $a\in A$, $f(a)=b$ and suppose there exists $f^{-1}:B\longrightarrow A$ that, given $b\in B$, $f^{-1}(b)=a$.

We define the identity function as follows:
$$Id_{A}:A\longrightarrow A$$
$$a\mapsto Id_{A}(a)= a$$
$$Id_{B}:B\longrightarrow B$$
$$b\mapsto Id_{B}(b)= b$$

If we define
$$f(f^{-1}):B\longrightarrow B$$
$$b\mapsto f(f^{-1}(b))=f(a)=b$$
And
$$f^{-1}(f):A\longrightarrow A$$
$$a\mapsto f^{-1}(f(a))=f^{-1}(b)=a$$

Therefore, the composition of inverses acts like $Id_{A}$ and like $Id_{B}$. As they are defined on the same domain, we have that:

$$f(f^{-1})=Id_{B}$$
$$f^{-1}(f)=Id_{A}$$


What is the intuition behind the law of quadratic reciprocity? Law of quadratic reciprocity states as follows:

Law of quadratic reciprocity — Let $p$ and $q$ be distinct odd prime numbers, and define the Legendre symbol as:
$$ \left(\frac {q}{p}\right)=\left\{\begin{array}{rl} 1 & \text{if } n^2\equiv q \pmod p \text{ for some integer } n, \\ -1 & \text{otherwise.} \end{array} \right.$$
Then:
$${\displaystyle \left({\frac {p}{q}}\right)\left({\frac {q}{p}}\right)=(-1)^{{\frac {p-1}{2}}{\frac {q-1}{2}}}.}$$

I have heard from class that there are hundreds of proof of this theorem. But the proof that I have learned in class is a very elementary one. As we know that many theorems in number theory have some very nice explanations using abstract algebra. Is there a proof of this theorem from the perspective of abstract algebra? And what is the intuition behind it? Thank you!

It would help if you told us what proof or proofs you already know.

From the perspective of algebraic number theory, the quadratic reciprocity law can be described using
the cyclotomic field $\mathbf Q(\zeta_p)$ and its unique quadratic subfield, which is $\mathbf Q(\sqrt{p^*})$
for $p^* = (-1)^{(p-1)/2}p$, where $p$ is an odd prime.

Write the quadratic reciprocity law (for two different odd primes $p$ and $q$) as
$$
\left(\frac{q}{p}\right) = \left(\frac{p^*}{q}\right).
$$
The intuition behind this formula is that, using the field extension $\mathbf Q(\zeta_p) \supset \mathbf Q(\sqrt{p^*})$ mentioned above, the two sides of this equation describe in different ways the Frobenius element associated to $q$ in ${\rm Gal}(\mathbf Q(\sqrt{p^*})/\mathbf Q)$, which as a quotient ${\rm Gal}(\mathbf Q(\zeta_p)/\mathbf Q) \cong (\mathbf Z/(p))^\times$ is $(\mathbf Z/(p))^\times$ modulo its squares.  For further details about this approach you need to learn algebraic number theory; this proof can't be described at an intuitive level. That's part of what makes the quadratic reciprocity law so mysterious.

The proof that best fits the intuition represented by this formula is Eisenstein's one.

Briefly, Eisenstein uses Gauss' lemma and finds that $(\frac{p}{q})(\frac{q}{p})$ equals $-1$ or $+1$  depending on the parity (odd/even) of the number of lattice points in the square with as boundary lines the $x$-axis, the $y$-axis, and the lines $x=\frac{p}{2}$ and $y=\frac{q}{2}$.

And that number is clearly equal to $\frac{p-1}{2}\cdot\frac{q-1}{2}$


How to remember the difference between "stars and bars" and multinomial coefficients? Explanation: It is easy for me to remember the difference between permutation coefficients and combination coefficients verbally: one says that the former

"is the number of ways of choosing $k$ things from $n$ options without replacement where order matters"

and that the latter

"is the number of ways of choosing $k$ things from $n$ options without replacement where order doesn't matter".

  Question: How can one quickly describe (verbally) the difference between multinomial coefficients and "stars and bars" (which will here henceforth be called multiset coefficients)? For which does order matter? For which does order not matter? For which is the choosing with/without replacement?

Attempt: Both involve a counting process with two considerations: (i) dividing objects of $n$ types into $i$ containers, (ii) each container with $k_i$ spots, so assigning $k = \sum_i k_i$ objects from the $n$ different types. So it seems like there might be some ambiguity, when one says "with/without replacement" or "order does/doesn't matter", is one referring to (i) or (ii)?

So it seems like there are at least four distinct possibilities:

Assign $n$ types of object into $i$ containers, where the order of the containers doesn't matter, with $k = \sum_i k_i$ spots total, where the order within each container doesn't matter.
Assign $n$ types of object into $i$ containers, where the order of the containers doesn't matter, with $k = \sum_i k_i$ spots total, where the order within each container does matter.
Assign $n$ types of object into $i$ containers, where the order of the containers does matter, with $k = \sum_i k_i$ spots total, where the order within each container doesn't matter.
Assign $n$ types of object into $i$ containers, where the order of the containers does matter, with $k = \sum_i k_i$ spots total, where the order within each container does matter.

IF this is true, then which of these options corresponds to multinomial coefficients, and which of these options corresponds to multiset coefficients?

Update: I think the difference is this: "multiset coefficients are the number of ways of distributing balls into boxes, where (i) the total number of balls is fixed, (ii) the number of balls in each container is not fixed, (iii) the balls are indistinguishable (their order doesn't matter, just the number in each container), (iv) the boxes are distinguishable (i.e. their order does matter)". (For (iv), e.g. $x^2 y \not= x y^2$.) While "multinomial coefficients are the number of ways of distributing balls into boxes, where (i) the total number of balls is fixed because (ii) the number of balls in each box is fixed, (iii) the balls are indistinguishable (hence the relationship to binomial coefficients), and (iv) the boxes are  distinguishable (but it doesn't matter in the case of binomials because the solution to the Diophantine equation $k_1 + k_2 = k$ is already determined once $k_1$ is chosen)".

So TL;DR: in both cases, the balls are indistinguishable (their order doesn't matter), the boxes are distinguishable (their order does matter), the total number of balls is fixed, but (a) for multiset coefficients, the number of balls in each box is not fixed in advance, while (b) for multinomial coefficients the number of balls in each box is fixed in advance.

Thus the difference is whether one fixes in advance how many balls are to be placed into each box.

A multinomial coefficient is often written in the form
$$
\binom{n}{k_1,k_2,\ldots,k_m}
$$
where $k_1+k_2+\cdots+k_m = n.$
Because of that last equation, the $n$ is redundant.
Binomial coefficients are technically multinomial coefficients as well,
but instead of writing
$\binom{n}{k_1,k_2}$ where $k_1+k_2 = n,$
we usually write either $\binom{n}{k_1}$ or $\binom{n}{k_2}.$
These are all mean exactly the same thing:
$$
\binom{k_1+k_2}{k_1,k_2}=\binom{k_1+k_2}{k_1}
=\binom{k_1+k_2}{k_2} = \frac{(k_1+k_2)!}{k_1!\,k_2!}.
$$

A multinomial coefficient can arise in a counting problem as follows:

You have $m$ containers, with sizes $k_1, k_2, \ldots, k_m$ respectively.
You have $n = k_1+k_2+\cdots+k_m$ distinguishable objects, which are just enough to fill every container when drawing from the $n$ objects without replacement.
The order of the containers matters.
The order of the objects within each container does not matter.

A multiset coefficient can arise as follows:

You have $n$ distinguishable objects.
You have a single container able to hold $k$ objects.
You draw objects from the $n$ objects with replacement, and put the copies of those objects into the container.
You continue putting copies of objects in the container until it is full.
The order of objects in the container does not matter.

A multiset coefficient can alternatively arise this way:

You have $k$ objects that are indistinguishable or that you choose not to distinguish.
You have $n$ containers, each with effectively infinite capacity (they never get full).
You draw the objects without replacement and put them in the containers.
The order of the containers matters.
The order of objects in a container cannot matter, because by assumption you cannot tell which object is which, only how many are in each container.

These are two quite different ways of modeling a multiset coefficient:
in one model, the number of objects is the first number of the coefficient, while in the other model the number of objects is the second number.
But there is a simple counting argument that shows both ways model the same total count:
each time you choose one of the $n$ objects in the first model to occupy one of the $k$ unordered (hence indistinguishable) places in the container, you choose one of the $n$ containers in the second model to hold one of the $k$ indistinguishable objects.
Just as the first model allows you to choose the same object as often as you want (until all spaces in the container are full),
the second model allows you to put objects in the the same container as many times as you like (until all objects have been placed).

While the term "multiset" is inspired by the first model, the second model is a frequent application of the "stars and bars" method.
It's useful to know how to apply the multiset coefficient to either model as needed.

Motivation: Describing the number of monomials of degree $d$ in the variables $x_0, \dots, x_n$ requires the use of a multiset coefficient, not a multinomial coefficient. When I forget why, I have difficulty explaining to myself why the multiset coefficient works and why the multinomial one doesn't.

Background: According to Wikipedia, they are distinct, but I am having difficulty focusing in on the key aspects which differentiate them, as well as the key aspects which make them similar.

  Unlike for binomial coefficients, there is no "multiset theorem" in which multiset coefficients would occur, and they should not be confused with the unrelated multinomial coefficients that occur in the multinomial theorem. (Source)

Multiset and multinomial coefficients are distinct, but even if they are not related to each other, they do appear to both be related to counting problems involving multisets.

  n multichoose k is given by the simple formula

  $$((n; k))=(n+k-1; k)=(n-1,k)!,$$
  where (n-1,k)! is a multinomial coefficient. (Source)

  The multinomial coefficients $$(n_1,n_2,...,n_k)!=\frac{((n_1+n_2+...+n_k)!)}{(n_1!n_2!...n_k!)} $$are the terms in the multinomial series expansion. In other words, the number of distinct permutations in a multiset of k distinct elements of multiplicity $n_i (1 \le i \le k)$ is $(n_1,...,n_k)!$ (Skiena 1990, p. 12). (Source)
  A multinomial coefficient is associated with each (finite) multiset taken from
  the set of natural numbers. Such a multi-set is given by a list $k_1, . . . , k_n,$ where
  numbers may be repeated, and where order does not matter... The total number of
  such multi-indices is $(\binom{n}{k})$.
  The multi-index N determines a multi-set $k_1, . . . , k_n$ taken from the natural numbers. This multi-set consists of the values of the function N. It has less information in it than the multi-index $N$, since it has lost the information about which elements of $B$ index the numbers. (Source)
  The number of multiset combinations equals the multinomial coefficient (n multichoose k). (Source)

My understanding currently is that the multinomial coefficient is the number of ways of assigning $n$ objects to $i$ containers with $k_i$ spots each and $k = \sum_i k_i$ spots total, where order matters.

This explanation makes sense to me inasmuch as it explains why the sum of all multinomial coefficients is $i^k = i^n$ when $k=n$. However, it doesn't make sense to me inasmuch as multinomial coefficients are supposed to be related to binomial coefficients, which are supposed to give the number of ways of choosing elements when order doesn't matter.

But if order doesn't matter for multinomial coefficients, it doesn't matter for multiset coefficients either, which doesn't leave a clear way to distinguish them:

  A k-combination with repetitions, or k-multicombination, or multisubset of size k from a set S is given by a sequence of k not necessarily distinct elements of S, where order is not taken into account. (Source)


Prove $\tan\big(\frac{\pi}{2} -\theta \big) = \cot \theta$ $$\tan\left(\frac{\pi}{2} -\theta \right) =  \cot \theta$$

I can prove this by changing into $\cos$ and $\sin$

But I want to know if it is possible to prove it using relation given below. If not why ?
$$\tan\left(\frac{\pi}{2} -\theta \right) =\frac{ \tan\frac{\pi}{2} -\tan\theta}{1+ \tan\frac{\pi}{2}\tan\theta} $$

No, it isn't possible with the identity in your question because $\tan \frac\pi2$ is not defined.  It isn't defined because $\cos \frac\pi 2 = 0$, and $\tan x = \frac{\sin x}{\cos x}$ in general.

Actually, it's exactly this fact ($\tan x = \frac{\sin x}{\cos x}$) that you'll want to use.  I'll get you started:

$$
  \tan\left(\frac{\pi}{2} - x\right) = \frac{\sin\left(\frac{\pi}{2} - x\right)}{\cos \left(\frac{\pi}{2} - x \right)}
   = \frac{\sin\frac\pi 2 \cos x - \cos \frac\pi 2 \sin x}{\cos \frac\pi 2 \cos x + \sin\frac \pi 2 \sin x}
$$

Of course it'd be simpler to use the cofunction identities $\sin(\frac \pi 2 - x) = \cos x$ and $\cos(\frac \pi 2 - x) = \sin x$, but I'm assuming that's not allowed (or you want something different) because your original question is basically asking for an explanation of the tangent cofunction identity.

Edit: Many other answers are mentioning limits, etc.  Technically those will work but I'm assuming you're currently a trig student, which means it's unlikely you've had formal exposure to limits, and especially to l'Hôpital's rule.

If instead of $$\ tan (\frac{\pi}{2} - x)$$ We take
$$ \lim_ {a \to \pi/2} \ tan(a-x)$$
Then $$\tan(a-x) = \frac{(\tan a  - \tan x)}{ (1+\tan (a)\tan(x))}$$
Divide by $\tan(a)$ and use $\frac{1}{\tan(a)}=0$


Is my implementation of FastICA right? I have written the following code to perform ICA on the mixed signals that were generated using a random mixing matrix. But the performance of the algorithm is not satisfactory. Did I do something wrong?

[x1, Fs1] = wavread('source2.wav');
[x2, Fs2] = wavread('source3.wav');
m = size(x1,1); % size of each signal
n = 2; % Number of sound sources
A = randn(n, n); %  Random 2 X 2 mixing matrix
x = A*[x1';x2']; % Mixed signals
mx = sum(x, 2)/m; % supposed to be the mean; E{x}
x = x - repmat(mx, 1, m); % x - E{x}
for i = 1:n                      % This step is done
    xi = x(i,:);                 % to whiten the data
    [E, D] = eig(xi*xi');        % (make variance one
    xi = E*sqrtm(inv(D))*E'*xi;  % and uncorrelate the data)
    x(i,:) = xi;
end
w = randn(n, 1); % Random weight vector
w = w/norm(w,2); % make 'w' a unit vector
w0 = randn(n, 1);
w0 = w0/norm(w0, 2);
while abs(abs(w0'*w)-1) > 0.000001
    w0 = w;
    w = x*G(w'*x)'/m - sum(DG(w'*x))*w/m; % This step is supposed to perform:
                                          % w = E{xg(w^{T}*x)} - E{g'(w^{T}*x)}w
    w = w/norm(w, 2);
end
sound(w'*x); % Supposed to be one of the original signals

The algorithm can be found here http://en.wikipedia.org/wiki/FastICA

In the above code G is the function $xe^{\frac{-x^2}{2}}$ (computes element wise) and DG is the derivative of G: $(1-x^2)e^{\frac{-x^2}{2}}$ (computes element wise).

Though the algorithm itself is correct, the part of the code that whitens the data is not correct. All I had to do to make the covariances of the data $1$ was the following.

c = cov(x');
sq = inv(sqrtm(c));
x = c*xx;

And I changed the function $G$ from the one in the question to $\tanh(x)$. It works like a charm now.

up vote
0
down vote
accepted
Though the algorithm itself is correct, the part of the code that whitens the data is not correct. All I had to do to make the covariances of the data 1 was the following.

$$c = \mbox{cov}(x');\\
sq = \mbox{inv}\left(\sqrt{m0(c)}\right);\\
x = c*xx;$$
And I changed the function $G$ from the one in the question to $\tanh(x)$. It works like a charm now.

Where do we paste the edited piece of code?


Bound for $e^{-\alpha x}$ For a part of my proof I need to establish that $e^{-\alpha x} \lt h(x)$, where $\alpha,x \gt 0, $ and $x,\alpha \in\mathbb{R}$. I thought for a while and couldn't find a function independent of $\alpha$ that fulfills my criteria. Any ideas?

Also note that:
$$e^{-\alpha x}=\sum_{n=0}^\infty\frac{(-1)^n\alpha^nx^n}{n!}$$
A polynomial of this expansion could be used, although take care to see that it is consistently greater as this is a series with alternating sign


Prove $x^4-18x^2+36x-27$ can never be a nonzero square rational when $x$ is rational Prove $x^4-18x^2+36x-27$ can never be a square rational (excluding 0), when x is rational

I have tried to use modulus, but didn't get anywhere, any help would be greatly appreciated.

The equation $y^2=x^4-18x^2+36x-27$ is birationally equivalent to the elliptic curve $w^2=z^3-432$, with $x=\frac{w-36}{2(z-12)}$. This curve has rank $0$ and torsion group $\mathbb Z/3\mathbb Z$, hence only two rational points $(z,w)=(12,\pm36)$. If we substitute these points into the formula for $x$, the positive $w$ gives the excluded $x=3$ solution and the negative $w$ incurs a division by zero. Hence $x^4-18x^2+36x-27$ can never be a nonzero rational square for rational $x$.

Say $x^4-18*x^2+36*x-27 = n^2$

The polynomial factors up

$(x-3)*(x^3+3*x^2-9*x+9) = n^2$

It means that if $x-3$ is a perfect square, then $x^3+3*x^2-9*x+9
$ is also a perfect square or both aren't perfect square

Just like $36=9×4=12×3$

If  $x-3$ is not a perfect square, then polynomial $x^3+3*x^2-9*x+9$ is not also a perfect, therefore it must factor up further, Just like $36=12×3=4×3×3$

But $x^3+3*x^2-9*x+9$ does not factor

It doesn't factor as supposed, it has to factor into $(x-3)*(x-a)^2$ in my definition, but there's no value of $a$ to fix this therefore $x^3+3*x^2-9*x+9$ does not factor to $(x-3)*(x-a)^2$

Say $x-3 = k^2$, $x = 3+k^2$

$(3+k^2)^3+3*(3+k^2)^2-9*(3+k^2)+9 = (n/k)^2$

$k^6+12*k^4+36*k^2+36 = (n/k)^2$

$k^2*(k^2+6)^2+36 = (n/k)^2$

$(k\cdot(k^2+6))^2 + 6^2 = (n/k)^2$

It's easy to see that no integer $k$ exist to that make this a perfect square

Therefore $x^4-18*x^2+36*x-27$ is not a perfect square


Proof of $\lim_{x\to\infty}1/x=0$ I'm trying to prove that the limit of $1/x$ approaching zero as $x$ approaches $\infty$.

My professor is asking that we use this definition: For all $\varepsilon>0$ there exists $m$ such that if $x>m$, then $∣f(x)−L∣<\varepsilon$.

I made it to the point of For all $\varepsilon>0$, if $x$ is greater than $\varepsilon$, then the absolute value of $1/x$ is less than epsilon. But I'm not sure if I went the right direction.

You wish to find an $m$ depending on $\varepsilon$ that makes the condition hold.

If $x>\frac1\varepsilon$ then
$$\left|\frac{1}{x}-0\right|=\frac1x < \frac{1}{\frac1\varepsilon} = \varepsilon, $$
thus you can choose $m = \frac1\varepsilon$.

The answer is a rephrasing of the original question: Is there a mathematical derivation to prove $\lim_{x \to \infty}\tfrac{1}{x} = 0$ without only assuming, guessing, or knowing the result and then confirming the result, $0$, using the definition of the limit, $L$?


Clock hands moving to take each others places. This is from 2012 Euclid. I was wondering if someone could point out the flaw in my reasoning here:
Let the minute hand point to $50+x$ minutes. Then the hour hand points to $45 + \dfrac{50 + x}{12}$ minutes because the hour hand moves at a rate 1/12 that of the minute hand. Since the hour hand will move from $45 + \dfrac{50 + x}{12}$ to $50+x$ the number of minutes it travels is
$50+x -45 - \dfrac{50 + x}{12} = \dfrac{11x + 10}{12}$
but since the minute hand travels to $45 + \dfrac{50+x}{12}$ and the hour travels to $50+x$ we know that
x = $\dfrac{45 + \dfrac{50+x}{12}}{12}$ and
x = $\dfrac{590}{143}$
Plugging this into $\dfrac{11x + 10}{12}$ we get 60/13 and we divide by 12 to convert minutes to hours, for a total of 5/13 hours. The correct answer is 12/13. I was hoping someone could point out the error in my solution, rather than providing a new one, as I have seen better ways to solve this using angles, but would like to see the error here.

Your calculation of $x$ is correct, and you are correct to conclude that the hour hand has moved $60/13$ minute marks from start to finish. But the hour hand moves $5$ minute marks per hour, so to obtain hours elapsed you should divide $60/13$ by $5$ (rather than divide by $12$), obtaining $12/13$ hours elapsed.

there are 60 minutes in hour.
So, the minute hand moves 12 times as quickly as the hour hand,so 60-x/x=12 or x=60/13 ---> eq.(1)
In one hour, the hour hand moves through 1/12 x 60 = 5. Since the hour hand is moving for t hours, then
5t = x
Substitute value of x from eq.1
5t = 60/13
t = 12/13


NP-complete impossible to solve in $O(n)$ NP-complete problems are likely to be unsolvable in polynomial time (although no one proved it yet). My question is, has anybody proved that they are unsolvable in $O(n^d)$ for some concrete small $d$? Say, has anybody proved that it's impossible to solve subset sum in $O(n)$?

Suppose that somebody had proved it for $O(n^d)$. Now assume $P = NP$ and fix some $NP$-complete problem that has a $O(n^r)$-time solution with $r > d$. If you pad the input of every instance, raising input length from $k$ to $k^{r/d}$ then the original algorithm works but the solution is produced in time $O((n^r)^{d/r}) = O(n^d)$, a contradiction.

In short: no. A proof that no $NP$-complete problem can have a $O(n^d)$-time solution is a proof that $P \neq NP$.

Edit: The link in zarathustra's comment frames your question in a way that avoids this kind of cheap padding trick.


Proving that a sequence is increasing Problem: A sequence $(a_n)$ is defined recursively as follows, where $0<\alpha\leqslant 2$:
$$
a_1=\alpha,\quad a_{n+1}=\frac{6(1+a_n)}{7+a_n}.
$$
Prove that this sequence is increasing and bounded above by $2$. What is its limit?

Ideas: How should I go about starting this proof? If the value of $a_1$ were given, I could show numerically and by induction that the sequence is increasing. But no exact value of $\alpha$ is given.

To prove the sequence is increasing consider the function

$$ f(x) = \frac{6(1+x)}{7+x}$$

and prove $f'(x)>0$ for $x>0$. To find the limit assume the limit is $a=\lim_{n\to \infty} a_n $ which gives

$$ a = \frac{6(1+a)}{7+a} $$

and solve for $a$.

I leave it for you to prove boundedness.


Is there any systematic way to factorise expressions of the form $ n^4+pn^2 +q$ Is there any systematic way to factorise expressions of the form $ n^4+pn^2 +q$

Expression like $n^4+n^2+1$ usually become helpful for telescopic type , when used in the form$(n^2+n+1)(n^2-n+1)$, but for an expression like $n^4+pn^2 +q= (n^2+an+b)(n^2+cn+d)$ systematic way to factorise it like above (if possible).

Yes, of course! Always there exist a way to factorize it.

If $p^2-4q\geq0,$ so we can use a way, which is a similar to the following:
$$x^4+4x^2+3=x^4+3x^2+x^2+3=(x^2+1)(x^2+3).$$
But for $p^2-4q<0$ there is a way, which is a similar to the following:
$$ x^4+x^2+25=x^4+10x^2+25-9x^2=(x^2-3x+5)(x^2+3x+5) $$

For $ p^2-4q<0 $ in the general case we obtain:
$$ x^4+px^2+q=(x^2+\sqrt{q})^2-2x^2\sqrt{q}+px^2=(x^2+\sqrt{q})^2-(2\sqrt{q}-p)x^2=$$
$$=\left(x^2-\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right)\left(x^2+\sqrt{2\sqrt{q}-p}x+\sqrt{q}\right) $$

This is a quadratic in $n^2,$ so you can always factorize it over the complex numbers.


How to define $x \in \mathbb N$ in the reals I just learned today about Tarski Seidenberg theorem which implies the decidability of the reals (only with the field operations). I also know about Gödel incompleteness theorem, which implies that Peano arithmetic is not decidable.

I was thinking then what is wrong with statements of the form
$$ x \text{ is a natural number} \wedge \phi(x) $$
Because then you would be able to encode statements about the naturals in the reals. Of course the problem is how to define something to be a natural number, and I was thinking by doing it like
$$x \text{ is a natural number} \iff x \ge 0 \wedge \sin(\pi x) = 0$$
And $\sin(x)$ can be define as an infinite series. I guess the problem in this case would be on how to define limits (I am not sure where is the problem here, but there has to be one).

My question is then, what extra thing would you need in order to define the statement $ x \text{ is a natural number}$? and why can't you define $\sin(x)$ in the reals using only the field operations?

Thanks,

If nothing else, in non-standard models of the reals, the statement "natural number" will not mean what you want. There is a non-standard model of the naturals embedded in every non-standard model of the reals.

Looking at an ultra power of $\mathbb{R}$, (the equivalence of) the sequence $n\mapsto n$ would satisfy any reasonable first order attempt to define "is a natural number" by Los's theorem. But this member of the ultra product fails to be a standard real.

First order simply isn't expressive enough to express "obvious" notions sometimes. Alternatively, our intuition about obvious is wrong and statements like "is a natural number" are more complicated than we naively realize.


Number of strongly-connected digraphs Is there a closed formula for the number of non-isomorphic
  strongly-connected digraphs on $n$ vertices?


Does unbounded derivative imply No Lipschitz? We are taught that bounded derivative implies lipschitz. Also, at times, derivative may not exist, yet lipschitz could hold.

I wonder if derivative exists but is unbounded, can we directly say that it is not lipschitz?

Assume $f$ is Lipschitz and differentiable at $I$ and let
$a\in I$.

$\exists K>0: $

$\forall x\in I - \{a\}$

$|\frac{f(x)-f(a)}{x-a}| \leq K$

then by passage to the limit when $x\to a$, we get

$|f'(a)|\leq K$ and $f'$ is bounded at $ I$.

$$f(x) = x \sin\left(\frac1x\right) \text{ on } (0,1)$$ is a Lipschitz function with unbounded derivative


Does $\lim_{n \to \infty} n \cos(n^2 x)$ exist for some $x$? (Michael Spivak "Calculus 3rd Edition" p.497) I am reading Michael Spivak "Calculus 3rd Edition".

On p.497 Spivak wrote as follows:

  if $$f_n(x) = \frac{1}{n} \sin(n^2 x),$$ then $\{f_n\}$ converges uniformly to the function $f(x) = 0$, but $$f_n^{'}(x) = n \cos(n^2 x),$$ and $\lim_{n \to \infty} n \cos(n^2 x)$ does not always exist(for example, it does not exist if $x = 0$).

Then,
does $\lim_{n \to \infty} n \cos(n^2 x)$  exist for some $x$?

If $\lim_{n\to\infty} n\cos(n^2x)$ exists then $\cos(n^2x)\to 0$ and hence $$n^2x=(2a_n+1)\frac{\pi}{2}+b_n$$ where $a_n\in\mathbb {Z}, b_n\to 0$. Replacing $n$ by $n-1$ and subtracting the resulting equation from the last equation we get $$(2n-1)x=c_n\pi+d_n$$ where $c_n\in\mathbb {Z}, d_n\to 0$. Applying the same argument again we see that $$2x=e_n\pi+f_n$$ where $e_n\in\mathbb {Z}, f_n\to 0$. Taking limits we see that $e_n\to 2x/\pi$. Since $e_n$ takes integer values it follows that the sequence $e_n$ ultimately becomes a constant, say $k$, and thus $x=k\pi/2$ for some integer $k$.

The case $k=0$ when $x=0$ is simplest one and has been mentioned by Spivak. For other values of $k$ the sequence $\cos(n^2x)$ oscillates or tends to $1$ (check this). And hence $n\cos(n^2x)$ never tends to a limit for any given value of $x$.

First, we observe that if $\cos n^2x \neq 0$, then $f_n$ does not have a limit, since $n \cos n^2x$ increases in magnitude.

Therefore, we look for cases where $\cos n^2 x$ can be zero so that the resulting sequence is the constant zero sequence, which will have a limit.

First, notice that if $n$ is odd, then $\cos n^2 x$ is zero for $x = \dfrac{\left( 2k + 1 \right) \pi}{2}$, where $k \in \mathbb{Z}$. However, for the same $x$, if $n$ is even, $\cos n^2x \neq 0$. Therefore, at this value of $x$, the sequence will not have a limit because it will have a seubsequence which increases in magnitude.

Similarly, if we try to make $\cos n^2x$ zero at even $n$, then the odd $n$'s create similar problems. Therefore, the sequence under consideration can never have a limit.


Number of 5 letter palindromes given 7 possible letters A 5 letter word is formed using some of the letters {a,b,h,i,p,r,s}.
How many of them will be palindromes?

Options:

$125$
$225$
$343$
$729$

My Approach:
Only the middle number can be arranged in $7$ ways. Rest all the numbers can be arranged like Whatever the first letter from right  should be last letter from left. Whatever the second letter from start should be last letter from left.
But i am not able to get to the solution.

Can anyone give me the hint to solve the problem.

**HINT:**Think about how many ways you have to choose the first letter, then the second, then the third and so on...

Pick the first center letter. At the right side you have a permutation of 7 letters in two i.e you 42 left and right letters for each center letter. So you will have 42x7 = 294 then add aa, bb, hh, ii,...for each center letter you will have 49 more. So the answer is c.


When is $32x+32$ a square number? I am trying to find out for what values for $x$ does the function $f(x)=32x+32$ return a square number? I found that this is the case for at least: $x \in \{64,1600,53824,1827904,62094400,2109381184\}$ but I can't see a pattern in those numbers.

Is there a formula which generates these values for $x$ instead of simply trying a lot of values?

$$32(x + 1) = y^2 \implies 32 \mid y^2 \implies 4 \mid y$$ so let $y = 4z$. $$2(x+1) = z^2 \implies 2 \mid z$$ so let $z = 2w$ $$x + 1 = 2w^2 $$ so given any $w$ we can solve for $x$.

So the set of solutions is
$$\{(2w^2 - 1, 8w) \mid w \in \Bbb Z\}$$

Solution for x such that 32x+32 is a square as follows:
 For integral solution,  x= (n2 -2)/2 such that x is in Natural number.(n is in natural number)


Diophantine equation $a^2+b^2=c^2+d^2$ I was reasonably certain I've seen this before, but I was wondering how to solve the Diophantine equation

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

I tried a web search and found nothing on this one.  I'm trying to avoid another library trip to a less than local library (maybe I should have taken better notes on that chapter...).

I'm not quite sure how to handle this one.  The only thing I can figure out with this equation, if I remember correctly, is that the sum on either side may only contain prime factors of 2 or odd primes congruent to 1 mod 4.  And if I don't want a and b equal to c and d, the sum can't be prime as I believe a prime congruent to 1 mod 4 can be represented as the sum of 2 squares in exactly one way.  But that doesn't give me any insight into actually solving this problem.

Essentially you ask for a parametrization of the quadric $V(a^2+b^2=c^2+d^2)$. There is a general geometric method how to do that, which you can find here. On page 13, this example is discussed: Solutions of $a^2+b^2=c^2+d^2$ are parametrized by

$(a,b,c,d) = (p r + q s , q r - p s , p r - q s , p s + q r),$

where $p,q,r,s$ are arbitrary. But one can also derive this via complex numbers (similar to the complex numbers solution of Pythagorean triples): Let $u = p + qi$, $v = r + si \in \mathbb{C}$. Then we have

$$|u \overline{v}| = |u| |\overline{v}| = |u| |v| = |u v|.$$

Square both sides, and compute the norms of the products explicitly. This gives you

$$(p r + q s)^2 + (q r - p s)^2 = (p r - q s)^2 + (p s + q r)^2.$$

Pick any point inside the square. The distances from the corners to that point have the relation  $a^2 +c^2 = b^2 +d^2$ where the lines to that point are labeled $a,b,c,d$ going clockwise.  Finding any point inside the unit square where all distances from the  corners to that point are rational is an unsolved problem.

If you did find such a point you would find integer solutions to those distances and the diophantine equation above. You also would have found eight pythagorean triple triangles that can be assembled into a square.


Show that the equation $\lambda - z - e^{-z} = 0$ has exactly one solution in the half plane. I know that this question has been asked before, but I'm not satisfied with the answers provided. I will post here my complete solution to this question, but I'm not very sure about my (a) and (c) parts.

Let $\lambda > 1$ and show that (a) the equation $\lambda - z - e^{-z} = 0$ has exactly one solution in the half plane $\{z: \Re(z) > 0\}$. (b) Show that this solution must be real. (c) What happens to the solution as $\lambda \to 1$?

Let f(z) = $\lambda - z - e^{-z}$

(a) The equation $f(z) = 0$ has exactly one solution in the half plane $\{z: \Re(z) > 0\}$.

Assume that $\exists z_0 \in G = \{z:\Re(z) > 0 \}$ such that $f(z_0) = 0$, we will show that this is the only solution in $G$.

$$f(z_0) = 0 \Rightarrow \lambda - z_0 = e^{-z_0} \Rightarrow |z_0 - \lambda| = e^{-\Re(z_0)}$$

But $Re(z_0) > 0$, thus

$$|z_0 - \lambda| < 1$$

Consider $h(z) = -e^{-z}, g(z) = \lambda - z$,

$$|h(z) + g(z)| = | \lambda - z - e^{-z} | \leq |e^{-z}| + |\lambda - z| = |h(z)| + |g(z)|$$

Hence, by Rouche's Theorem

$$Z_h + P_h = Z_g + P_g$$

But $P_h = P_g = 0$. Since $g$ has only one zero in $|z - \lambda| < 1$, $h + g = f(z)$ has only one zero also. $\square$

(b) Show that the solution is real

Consider the real function

$$f(x) = \lambda - x - e^{-x}$$

Then $f(0) = \lambda - 1 > 0$ and $f(\lambda) = -e^{-\lambda} < 0$, thus by the Intermediate value theorem,

$$\exists z_0 \in (0,\lambda), \mbox{ such that } f(z_0) = 0.$$

But this solution is unique by part (a), hence $f$ has a unique real solution in $G$.$\square$

(c) What happen to the solution $\lambda \to 1$?

$$
\begin{align*}
\lim_{\lambda \to 1}f(z_0) & = \lim_{\lambda \to 1}0 \\
1 - z_0 - e^{-z_0} & = 0 \\
z_0 + e^{-z_0} = 1
\end{align*}
$$

Hence, $z_0 \to 0$ as $\lambda \to 1$. $\square$

My problem is that I can not see how we apply here the rouche theorem for make the claim about the roots of $f$ when we only consider $h$ and $g$ and not $f$. That point is not very clear to me.

[For completeness sake]

Since $\lambda > 1$, note that $ \gamma \doteq \partial B(\lambda,1) $ is in the right semi-plane.

Consider the functions $f(z) = e^z + z - \lambda $ and $g(z) = \lambda - z$, as above.

Notice that at $\{\gamma\}$ we have:

$$
|f(z) + g(z)| = |e^z| < 1 = |g(z)| < |f(z)| + |g(z)|
$$

then by Rouché, since $f$ and $g$ don't have any poles, $f$ must have exactly one zero  $\beta \in B(\lambda,1)$, just like $g$.

On the other hand, let $\alpha $ in the right semi-plane be a root of $f$,
$$
|\lambda - \alpha| =
|e^\alpha| < 1
\implies
\alpha \in B(\lambda,1)
\implies
\alpha = \beta
$$

then $\alpha$ is unique.

Part (b) of this problem was not correct. I figured out how to solve my own problem by doing the following:

Consider $g(z) = -\lambda + z$,

$$
|f(z) + g(z)| = | \lambda - z - e^{-z} + -\lambda + z | = |-e^{-z}|
$$

Then

$$
|f(z)| + |g(z)| = |\lambda - z - e^{-z}| + |-e^{-z}|
$$

Thus,

$$
|f(z) + g(z)| = |-e^{-z}| < |f(z)| + |g(z)| = |f(z)| + |e^{-z}|
$$

Hence, by Rouche's Theorem

$$
Z_f + P_f = Z_g + P_g
$$

But $P_f = P_g = 0$. Since $g$ has only one zero in $|z - \lambda| < 1$, then $f(z)$ has only one zero also.$\square$


Spivak Calculus on Manifolds - Problem 1.22 Problem 1.22 - Spivak's Calculus on manifolds
If $U$ is open $C \subset U$ is compact, show that there is a compact set $D$  such that $C \subset interior \ D$ and $D \subset U$.
observation: the metric space is $\mathbb{R}^n$.

My attempt:
$U$ is an open set, so exists $\delta_x > 0$ such that $B(x,\delta_x) \subset U$ for each $x \in U$, therefore the collection $\{ B(x,\delta_x) \}$ is an open cover for $C$ since $U = \bigcup_{x \in U} B(x,\delta_x)$ and $C \subset U$ then exists a finite subcollection $\{ B(x_i,\delta_{x_i}) \ ; \ i = 1, \cdots, n \}$ that covers $C$ by compactness of $C$. I suppose that my candidate of $D$ is $\overline{ \bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i}) }$ since the set $\bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i})$ is open and $interior \ D = \bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i})$ and $C \subset \ interior \ D$.
I would like to know if I'm right of thinking that $D = \overline{ \bigcup_{i = 1}^{i = n} B(x_i,\delta_{x_i}) }$ and I would like to know how to prove that $D$ is compact and that $D \subset U$.
Thanks in advance!

You are almost there, just replace $B(x_i,\delta_{x_i})$ by $B(x_i,\delta_{x_i}/2)$ and take $D=\bigcup_{i=1}^{i=n}\overline{B(x_i,\delta_{x_i}/2)}$.

$\overline{B(x_i,\delta_{x_i}/2)}\subset B(x_i,\delta_{x_i}) $ implies $D\subset U$. $D$ is compact since a closed ball in $R^n$ is compact and a finite union of compact subsets is compact since it is closed and bounded.

In the first line, you claim that there exists $d > 0$ such that $B(x,d)$ is contained in $U$ for all $x$. However, that may not be the case, as the d depends on x. However, if you use $d_x$ and $d_x/2$ in the proof then this doesn't really affect anything.


Linear Algebra Basis Trick Let $V$ be a finite dimensional vector space over a field $F$. More generally, we can consider a free module $M$ over a commutative ring $A$, with rank $n$.

Let $(m_1,...,m_n)$ be a basis for $M/A$ and $(m_1',m_2',...,m_n')$ be another basis.

Any linear map $L:M\to M$ over $A$, has a corresponding matrix with respect to $m_i$ and $m_i'$. With respect to $m_i$ the matrix is given by $(a_{ij})$ where $Lb_j = \sum_i a_{ij}b_i$ and similarly for $(a'_{ij})$. We also have another matrix $(u_{ij})$, called the change-of-basis, from $m_i$ to $m_i'$, given by $m_j = \sum_i u_{ij}m_j'$.

I always forget the relationship between $(a_{ij})$ and $(a'_{ij})$. Of course, I can derive it, but it is annoying when one has to do it again and again. What is the right way to conjugate? Is there some easy-to-remember technique, most preferably, using a commutative diagram, that we can use? I really hate it when my train of thought gets derailed trying to figure out which matrix goes where.

As an additional question. How about those computations including the bilinear form matrix and its change-of-basis and those including discriminants of number fields with respect to a basis? Is there a generally convenient way of looking at this?

The relation that uses the coefficients of the matrix  $(a_{ij})$ is wrong. Don't forget matrix multiplication is done multiplying $row$ of the first matrix by columns of the second matrix. So the relation is:
$$Lb_i=\sum_{j=1}^n a_{ij}b_j$$

The change of basis matrix is also wrong: its column-vectors are made up of the coordinates of the new basis: $(m'_1,\dots,m'_n)$ in the old basis: $(m_1,\dots,m_n)$.

But it allows to compute ‘old’  coordinates (wrt the old basis) of a vector from their ‘new’ cordinates. So if we denote $X$ the column vector of coordinates in the old basis, $X'$ the column vector of the same vector in the new basis, $Y$ the column vector of old coordinates of the image of $X$ by $L$, and similarly for $Y'$, we have the relations ($A$ is the matrix of the transformation $L$ in the old basis, $A'$ its matrix in the new basis):

$$Y=AX,\qquad X= PX',\qquad Y=PY',$$
hence $$PY'=A(PX')\iff Y'=(P^{-1}AP)X'$$
so that $\,\,A'=P^{-1}AP$.

Here is a reminder in the form of a commutative diagram: both horizontal arrows are $f$, and both vertical arrows are the identity map, in the relevant bases. Just remember the change of basis from $\mathcal B$ to $\mathcal B'$ matrix is but the matrix of the identity map from $(E,\mathcal B')$ to  $(E,\mathcal B)$ :

There's some notation, which I think I learned from Lang's Algebra, that seems to make these things easier. If $f\colon V \to W$ is linear and we have bases $\mathbf v$ and $\mathbf w$ then ${}_{\mathbf w}[f]_{\mathbf v}$ is the matrix with respect to these. We have the property that ${}_{\mathbf w}[g]_{\mathbf v} \cdot {}_{\mathbf v}[f]_{\mathbf u} = {}_{\mathbf w}[g \circ f]_{\mathbf u}$.

Now, if I have two bases $\mathbf v$ and $\mathbf v'$ for $V$ then your  change of basis matrix from the first to the second R is just ${}_{\mathbf v'}[\operatorname{id}_V]_{\mathbf v}$. And now you can use  the composition rule.


Derivative of $(-1)^x$, several cases How would you differentiate the function $(-1)^x$ w.r.t. $x$.
Is the formula for $a^x$ valid here? Wolfram says that the derivative of $(-1)^x$ is $iπ(-1)^x$. So, does this mean than $ln(-1)=iπ$ ?

Another thing that intrigued me was that its domain and range are given to be $\emptyset$. Is it correct?(or does Wolfram Mathematica wants to say that a real domain and range do not exist?)

Update:

There arise 2 cases

1) $f_1 : \mathbb{R} \rightarrow \mathbb{R}$

$f_1(x)=(-1)^x$

2) $f_2 : \mathbb{C} \rightarrow \mathbb{C}$

$f_2(x)=(-1)^x$

Then as per the comments, $f_1$ is not differentiable and its domain and range cannot be specified, while $f_2$ is a differentiable function.(Right?)

Since it seems you are insisting in asking for the domain of $(-1)^x$ for $x \in \mathbb R$, then
note that
$$
\left( { - 1} \right)^{\,x}  = \left( { - 1} \right)^{\,\left\lfloor x \right\rfloor  + \left\{ x \right\}}  = \left( { - 1} \right)^{\,\left\lfloor x \right\rfloor } \left( { - 1} \right)^{\,\left\{ x \right\}}
$$
$(-1)$ elevated at the fractional part of $x$ is not defined in the reals, unless of course $\left\{ x \right\}$ be null
$$
\left( { - 1} \right)^{\,\left\{ x \right\}} \text{n}\text{.}\,\text{d}\text{.}\quad \left| {\;0 < \left\{ x \right\}\; \wedge \;x \in R} \right.
$$

so we can write
$$
y = \left( { - 1} \right)^{\,x} \quad \left| {\;x \in R} \right.\quad  = \left\{ {\begin{array}{*{20}c}
   {\left( { - 1} \right)^{\,x}  = \left( { - 1} \right)^{\,\bmod \left( {x,2} \right)} } & { \Rightarrow \quad y \in \left\{ { - 1,1} \right\}} & {\left| {\;x \in Z\,} \right.}  \\
   {n.\,d.} & { \Rightarrow \quad y \in \emptyset } & {\left| {\;x \notin Z} \right.}  \\
 \end{array} } \right.
$$

$$f(z) = (-1)^{z} = \mathrm{e}^{z\ln(-1)}$$

\begin{align}
\ln(-1) &= \ln|-1| + i\mathrm{Arg}(-1) + i2\pi n \\
&= 0 + i\pi +i2\pi n \\
&= i\pi (2n+1)
\end{align}
where

$n \in \mathbb{Z}$
We designate the principal branch of the logarithm as $-\pi \lt \mathrm{Arg}(z) \le \pi$
On the principal branch, $n=0$ and $\ln(-1) = \mathrm{Ln}(-1) = i\pi$

Thus
$$\frac{\mathrm{d}f}{\mathrm{d}z} = \ln(-1)\,(-1)^{z} = (-1)^{z}[i\pi (2n+1)]$$

On the principal branch
$$\frac{\mathrm{d}f}{\mathrm{d}z} = i\pi (-1)^{z}$$


Evaluate the limit $\lim\limits_{n\to0}\frac{(x)+(2x)+\cdots (nx)}{n^2}$ Find the limit of $\lim_{n\rightarrow ~0}\frac{(x)+(2x)+\cdots (nx)}{n^2}$, where, $(x)=x-[x]$ and $[x] $ is the greatest integer function(the fractional part function).

I feel, as $n \rightarrow 0$ this limit  goes to infinity, but the options given are $x~,~x/2,~x/3,~x/4$.

How this is happening, I double checked the question paper, in question $n$. is tending to 0 only not to $\infty$.

I found a similar question here

This was too long for a comment, hopefully this helps:

Using $1+2+\dots+n=\frac{n(n+1)}{2}$ we find that:
\begin{align*}
\lim_{n\rightarrow\infty} \frac{(x)+(2x)+\dots+(nx)}{n^2}
&=\lim_{n\rightarrow\infty} \frac{x-[x]+2x-[2x]+\dots+nx-[nx]}{n^2}\\
&=\lim_{n\rightarrow\infty} \frac{x(1+2+\dots+n)}{n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
&=x\lim_{n\rightarrow\infty} \frac{n(n+1)}{2n^2}-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
&=\frac x2-\lim_{n\rightarrow\infty} \frac{[x]+[2x]+\dots+[nx]}{n^2}\\
\end{align*}

hint

$$\lim{n\rightarrow \infty} \frac{(x) [1+2+3 +. . . + n]}{n^2}$$
 As$ 1+2+3+. . .+n = \frac{n(n-1)}{2}$
So
$$\lim{ n \rightarrow \infty} \frac{1-\frac{1}{n}}{1} x$$
 Answer is $x$


Solve $n * \lceil{\frac{R}{T}}\rceil - \lceil{\frac{R*n}{T} - \frac{x*n}{T}}\rceil = \frac{x * n}{C}$ for x and n, $n \in Z^+$ and $x,R,T,C \in R^+$ I want to find the minimum value of n which satisfy given equation.

(Also, not stated in title but given is that $C < T$)

So far, I have been able to find the following properties:

$$\frac{x * n}{C} \in Z^+$$
$$x \leqslant \frac{\lceil{\frac{R}{T}}\rceil - \frac{R}{T}}{\frac{1}{C} - \frac{1}{T}}$$

Reasoning for the first observation is that since left hand side of the equation in title is all integers, right hand side too needs to be integer

The second observation comes from the following:

$$\frac{R*n}{T} - \frac{x*n}{T} \leqslant \lceil{\frac{R*n}{T} - \frac{x*n}{T}}\rceil$$

So

$$ n * \lceil{\frac{R}{T}}\rceil - \lceil{\frac{R*n}{T} - \frac{x*n}{T}}\rceil = \frac{x * n}{C} \leqslant n * \lceil{\frac{R}{T}}\rceil - (\frac{R*n}{T} - \frac{x*n}{T} ) \implies$$

$$ x*n(\frac{1}{C} - \frac{1}{T}) \leqslant n *(\lceil{\frac{R}{T}}\rceil - \frac{R}{T})$$

But I haven't been able to find any explicit constraints on n. What is the minimum bound on n?

Also, I suspect but can't prove, that if there exists a value of n, then there is no upper bound on n.

For context, I have asked a similar (but not exactly the same) question here: Solve $\lceil x\rceil-\frac{\lceil nx\rceil}n\geqslant y$ for $n$, where $n \in\mathbb Z^+, x \in\mathbb R^+, y \in\mathbb R^+$

Also, I need to solve this equation in a computer program. So an iterative solution also works.

Trivial but perhaps not efficient: Iterate over $n=1,2,\dots$ until you get a solution. For each $n$, substitute it (so that the only variable is $x$) and try to solve equation for $x$. Note that the left side is decreasing in $x$ and right side is increasing in $x$. Something "primitive" like bisection search should work.


Euclid: What is the difference between a 'surface' and a plane 'surface'? I've begun to study Euclid's Elements and i've a few questions regarding the difference between a surface and plane surface.

A surface is said to be "that which has length and breadth only", it then goes on to say "a plane surface is a surface which lies evenly with the straight lines on itself".

What is the difference between the two? What exactly is a plane surface? Can a plane surface be curved, or is it only flat?

I apologise for being a pedant, these definitions should be straight forward but I wish they could be a bit more specific, I want a good understanding of them before I move on.

The first point to make is that Euclid's definitions are not definitions in the modern sense, so that this kind of confusion is common. A surface, for Euclid, is roughly a two-manifold embedded in 3-dimensional space, e.g. the standard 2-sphere or one of its hemispheres, or the graph of certain functions from the plane to the line. A plane surface is just a surface that happens to lie in a plane, e.g. the unit disk. One more modern way to think about "lies evenly with the straight lines on itself" is that the tangent lines to a plane surface lie within the surface, as does not occur in general.

In differential geometry a 2-D surface is like a sphere which has length and breadth, Gauss curvature K is non-zero. It is doubly curved.

A flat surface has zero K and is a part of a plane or developable to it, at least one of two radii of curvature is infinity.

If you try to flatten a Pringles saddle shaped chip, circumferential lines tend to overlap before they break up. On the other hand when you step on an egg shell crushing it, it splits the boundary circumferentially.

Such differential deformation behavior is at the back of intuitive physical perception of spatial embedding of positive, negative and zero double curvature of surfaces.. while engrossed in the formalism of their mathematical treatment.


Can an irrational number raised to an irrational power be rational? Can an irrational number raised to an irrational power be rational?

If it can be rational, how can one prove it?

There is a classic example here. Consider $A=\sqrt{2}^\sqrt{2}$. Then $A$ is either rational or irrational. If it is irrational, then we have $A^\sqrt{2}=\sqrt{2}^2=2$.

Let me expand orangeskid's answer, both because I think it teaches us something useful, and it might be the easiest elementary proof for this question.
The proof that $ \sqrt{2} $ is irrational is well-known, so I will not repeat it here.
But there's a proof just as simple showing that $ \log 3 / \log 2 $ is irrational.  Suppose on contrary that $ \log 3 / \log 2 = p / q $ where p and q are integers.  Since $ 0 < \log 3 / \log 2 $, we can choose $ p $ and $ q $ both as positive integers.  The equality then rearranges to $ 3^q = 2^p $.  But here, the left hand side is odd and the right hand side is even, so we get a contradiction.
That gives a positive answer to the original question:
$$
\big(\sqrt{2}\big)^{2 \log 3 / \log 2} = 3
$$
Thanks to Rand al'Thor, who mentioned this problem in SE chat and thus inspired this answer.