Non-inductive, not combinatorial proof of $\sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ I've seen the identity $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ used here recently.

I checked for proofs here http://www.proofwiki.org/wiki/Sum_of_Squares_of_Binomial_Coefficients

I couldn't have figured out the combinatorial proof by myself, and the inductive proof assumes you already know the answer...

So my question is : do you know how to prove directly through computation that $\displaystyle \sum_{i \mathop = 0}^n \binom n i^2 = \binom {2 n} n$ ?

Consider the identity $(1+x)^{2n}=(1+x)^n\cdot(1+x)^n$.  By the binomial theorem we have $\displaystyle(1+x)^n=\sum_{k=0}^n{n\choose k} x^k$, so multiplying out we compute the right hand side as $\displaystyle(1+x)^n\cdot(1+x)^n = \sum_{k=0}^{2n}\left(\sum_{j=0}^k{n\choose j}{n\choose k-j}x^k\right)$.  But the LHS is just $\displaystyle(1+x)^{2n} = \sum_{k=0}^{2n}{2n\choose k}x^k$; equating coefficients of $x^n$ we get $\displaystyle{2n\choose n}=\sum_{j=0}^n {n\choose j}{n\choose n-j}$.  Finally, using the identity ${n\choose j}={n\choose n-j}$ gives the desired result.

Note that we can divide the $2n$ objects in $2$ groups including $n$ elements each.

Then to choose $n$ elements out of $2n$ we can choose $0$ objects in the first group and $n$ in the second, or $1$ in the first and $n-1$ in the second and so on, i.e $$\binom{2n}{n} = \sum_{i = 0}^{2n}\binom{n}{i}\binom{n}{n-i} =\sum_{i = 0}^{2n}{\binom{n}{i}}^2 $$


Identify the set of strings of odd length over {a, b} that contain the substring bb I need identify the set of possible strings of odd length over $\sum$ (the alphabet) {a, b} that contain the substring bb.

This is new material and a new concept to me and would really appreciate any perspective on how to go about solving this (with an explanation of course).

The only ways I see possible of solving this is through a mathematical proof by induction approach? Or by generating a regular-expression that defines this language (which is also something I feel is a bit ambiguous).

HINT: You could start by finding a regular expression for the strings that contain the substring $bb$, irrespective of their length: the simplest is

$$(a+b)^*bb(a+b)^*\;.\tag{1}$$

Each of the subexpressions $(a+b)^*$ matches the set of all strings over $\{a,b\}$, so $(1)$ matches anything of the form $ubbv$ with $u,v\in\Sigma^*$, i.e., any string that has the substring $bb$ in it somewhere.

But we want only the strings $ubbv$ of odd length. Now $|uv|=|ubbv|-2$, so $ubbv$ has odd length if and only if $uv$ has odd length. And $|uv|=|u|+|v|$, so we want to generate all pairs of strings $u$ and $v$ such that $|u|+|v|$ is odd. That happens precisely when one of $|u|$ and $|v|$ is odd and the other is even.

Explain why the regular expression $$(aa+ab+ba+bb)^*\tag{2}$$ matches precisely the strings over $\Sigma$ that have even length.
Find a regular expression that matches precisely the strings over $\Sigma$ that have odd length; you may find it helpful to start with $(2)$ and modify or add to it.
Use the previous results to get a regular expression that matches precisely the words $ubbv\in\Sigma^*$ such that $|u|$ is odd and $|v|$ is even.
Do the same for words $ubbv$ such that $|u|$ is even and $|v|$ is odd.

Now put the pieces together to get a regular expression that matches precisely those words over $\Sigma$ that contain the substring $bb$ and have odd length.

A regular expression to generate strings of odd lengths:

         (a+b)+a|b(aa+ab+ba+bb)*


If $x, y, z\in\mathbb R^+ $ and $x^3+y^3=z^3,$ then prove that $x^2+y^2-z^2>6(z-x) (z-y). $ I made several unsuccessful attempts. Still couldn't think of a proper way to prove the inequality. Please suggest how to approach this problem. Thanks in advance.
EDIT 1. My approach (that I was talking about):
Given: $z^3=x^3+y^3.$
We have to prove:
$x^2+y^2-z^2>6(z-x) (z-y)$ i.e., $\underbrace{(z^2+zx+x^2) (z^2+zy+y^2) (x^2+y^2-z^2)}_{=E\text{ (say)}}>6(z^3-x^3) (z^3-y^3)=6x^3y^3.$
(Here one thing which I noticed is that $(x^2+y^2-z^2)>0,$ since each of the terms on the LHS except this one is positive and $6x^3y^3$ is also positive for $x, y, z>0.$)
Using AM $\ge$ GM, we have:
$E\ge 3zx\cdot3zy\cdot(x^2+y^2-z^2)\ge 9xyz^2(2xy-z^2).$
From here I couldn't think of a proper way to prove $E>6x^3y^3.$ But I'm still working on it. At present I'm trying to manipulate the expression $9xyz^2(2xy-z^2)$ to get the job done. If I find something useful I'll update here.

Hint: eliminate $z$ from the system
$$ \eqalign{x^3 + y^3 &= z^3 \cr
      x^2 + y^2 - z^2 &= 6 (z-x)(z-y)\cr} $$
and show the only real solutions of the resulting equation are $x=0$ and $y=0$.


Argue that the sequence converges Let $$a_n=n^2\cos(1/n)-n^2$$
Show that the sequence converges.
Now, I know how to use the formal definition of convergence but I am looking for simpler methods (i.e the tests for series). I found it to converge to $-1/2$ but I need some method to show that it actually converges. Same problem applies to the sequence (not series)
$$\frac{n^3}{n^2}-\frac{n^3}{(n+1)^2} $$
I know the process of finding the value as $n$ shoots off to infinity for this sequence would just be write it all under 1 fraction and divide by the biggest power of $n$. But I run into the same problem of how to argue that it does indeed converge.

They are not similar problems. For
$$
\frac{n^3}{n^2}-\frac{n^3}{(n+1)^2},
$$
you just do a tiny bit of elementary manipulation to write it as
$$
\frac{n(n+1)^2-n^3}{(n+1)^2}=\frac{2n^2+n}{(n+1)^2}=\frac{2+1/n}{(1+1/n)^2}\xrightarrow[n\to\infty]{}2
$$
For the cosine, you need to use some property specific to the cosine. If you know that $\cos(t)=1-\frac{t^2}2+o(t^4)$, you get
$$
n^2\cos\frac1n-n^2=\frac12+o(\frac1{n^2})\xrightarrow[n\to\infty]{}0.
$$
If for instance you know that $\lim_{t\to0}\frac{\sin t}t=1$, you can do, from $\cos^2t-1=(\cos t -1)(\cos t+1)$,
$$
\cos t -1 =\frac{\cos^2t-1}{\cos t+1}=-\frac{\sin^2t}{\cos t+1}.
$$
Then
\begin{align}
n^2\cos\frac1n-n^2&=n^2(\cos\frac1n-1)=-n^2\,\frac{\sin^2\frac1n}{\cos\frac1n+1}\\[0.3cm]
&=-\frac1{\cos\frac1n+1}\,\Bigg(\frac{\sin\frac1n}{\frac1n}\Bigg)^2\xrightarrow[n\to\infty]{}-\frac12\times1=-\frac12.
\end{align}


Dimension of the vector space of homogeneous polynomials Let $k[X_0, X_1, \ldots, X_n]_d$, or briefly $k[X]_d$, be the $k$-vector space whose elements are the zero polynomial and homogeneous polynomials of degree $d\geq 1$. I found the following formula for the dimension
$$dim(k[X]_d)= \binom{n+d} {n}$$

but in my book there is no justification for this equality. Could someone explain me, possibly in an intuitive way, why that binomial coefficient is the dimension of that vector space?

Added later: Except the formula is correct, as shown in the comments below; I had mis-read it; when $d = 2$ and you have 3 variables, $n = 2$, since the variable indexing starts at zero. And 4-choose-2 is in fact six, as expected.

[what follows is incorrect]

A nice basis for that space consists of all monomials in the $n$ variables with total degree $d$.

Wait...what about degree $d = 2$ and $n = 3$ variables. Listing the basis elements, I see
$$
x^2, y^2, z^2, xy, xz, yz,
$$
which is only $6$ dimensions, but 5 choose 3 is 10. Seems as if there might be a mistake in your formula, or perhaps I'm not understanding how it's supposed to be applied. (See below for resolution of this apparent contradiction.)

I think (or thought) perhaps your formula is incorrect.

A degree 2, also known as quadratic, polynomial in the 1 variable  x  is of the form
                      a*x^2 + b*x^2 + c

for some numbers  a,b,c . The polynomial is determined by the 3 coefficients  a,b,c , and different choices of  (a,b,c)  result in different polynomials.
In linear algebraic terms, the space of degree 2 polynomials in 1 variable is of dimension 3 since it consists of all linear combinations of 3 linearly independent vectors  x2,x,  and  1 .
A degree 2 polynomial in 2 variables  x1,x2  is of the form
             $a*x_1^2 + b*x_2^2 + c*x_1*x_2 + d*x_1+ e*x_2 + f$

for some numbers  a,b,c,d,e,f . Different choices of  (a,b,c,d,e,f)  result in different polynomials.
In linear algebraic terms, the space of degree 2 polynomials in 2 variables is of dimension 6 since it consists of all linear combinations of 6 linearly independent vectors x_1^2,x_2^2,x_1x_2,x_1,x_2, and 1.


Proving that $\lim_{n \to \infty} \int_0^1 f(x) \sin(nx) \, dx = 0$ if $f : \mathbb R \to \mathbb R$ is continuous Specifically I would like to know how one can justify the following equality
$$\begin{align}
&2 \int_0^1 f(x) \sin(nx) \, dx \\
= &\int_0^\frac{\pi}{n}f(x)\sin(nx) \, dx - \int_1^{1+\frac{\pi}n} f(x)\sin(nx) \, dx + \int_0^1 \left(f(x)-f\left(x-\frac {\pi}n \right) \right) \sin(nx) \, dx
\end{align}
$$
knowing that (as $\sin(nx)$ has a period of $\frac{2\pi}n$)
$$
\int_0^1 f(x) \sin(nx) \, dx = -\int_{\frac{\pi}n}^{1+\frac{\pi}n} f\left(x-\frac {\pi}n \right)  \sin(nx) \, dx
$$
for all continuous functions $f: \mathbb R \to \mathbb R$?

The equality in question is taken from page 1 of this PDF file and is used to prove that
$$
\lim_{n \to \infty} \int_0^1 f(x) \sin(nx) \, dx = 0
$$
if $f : \mathbb R \to \mathbb R$ is continuous.

I am already aware of the Riemann Lebesgue Lemma, but I wanted to understand precisely the justification of the equality in question. The remainder of the proof follows easily for me.

The equality is incorrect.

Let $f(x) = x$.  Then we get

$$\begin{align}2 \int_0^1 x \sin nx \, dx &= \frac{2 \sin n}{n^2}-\frac{\cos n}{n} \\ &\neq \frac{2 \sin n}{n^2} +\frac{2\pi}{n^2} - \frac{2 \cos n}{n} - \frac{2 \pi \cos n}{n^2}\\ &=  \int_0^{\pi/n} x \sin nx \, dx - \int_{1}^{1 + \pi/n} x \sin nx \, dx + \int_0^1 [x - (x -\pi/n)] \sin nx \, dx\end{align}$$

The correct result is

$$\begin{align} 2 \int_0^1 f(x) \sin nx \, dx  &= \int_0^\frac{\pi}{n}f\left(x-\frac {\pi}n \right)\sin(nx) \, dx - \int_1^{1+\frac{\pi}n} f\left(x-\frac {\pi}n \right)\sin(nx) \, dx \\ &+ \int_0^1 \left(f(x)-f\left(x-\frac {\pi}n \right) \right) \sin(nx) \, dx \end{align}
$$

This follows easily by splitting the integral over $[\pi/n, 1+\pi/n]$ on the RHS of the first line of the solution in the attached PDF into integrals  over $[\pi/n,1]$ and $[1,1 + \pi/n]$ and then adding and subtracting the integral over $[0,\pi/n]$.

By Heine-Cantor, $f$ is uniformly continuous.

Fix $\varepsilon > 0$. Let $\delta > 0$ such that $|x-y| < \delta \implies |f(x) -f(y)| < \varepsilon$. Let $N \in \Bbb N$ such that $\frac {2 \pi} N < \delta$. For any $n > N$:

$$\begin{array}{cl}
&\displaystyle \left| \int_0^1 f(x) \sin(nx) \ \mathrm dx \right| \\
\le& \displaystyle \sum_{k=0}^m \left| \int_{2\pi k/n}^{2\pi (k+1)/n} f(x) \sin(nx) \ \mathrm dx \right| + \frac{2\pi}n M \\
\le& \displaystyle \sum_{k=0}^m \frac{2\pi \varepsilon}n + \frac{2\pi}n M \\
\le& \varepsilon + \dfrac{2\pi}n M
\end{array}$$

where $m = \left\lfloor\dfrac{n}{2\pi}\right\rfloor - 1$.

Now use standard tricks in analysis.


Calculus Word Problem: Several Variables Demand for a certain kind of SUV obeys the following equation,

$$D(x, y) = 21000 - \frac {\sqrt{x}}{2} - 11(0.3y-10)^{3/2}$$

where x is the price per car in dollars, y is the cost of gasoline per litre, and D is the number of cars.

Suppose that the price of the car and the price of gasoline t years from now obey the following equations:

$$x  =  55,200 + 100t ~~~~and~~~~      y = 136 + 10 \sqrt{t}$$

What will the rate of change of the demand be (with respect to time) 5 years from now?

I would really appreciate getting some help answering this question. I truly don't know where to start with this question never mind the solution. What practice or formula should I be using to answer this question?

New info

First I plug t= 5 into both equations. so

$x= 55700$

$y=158$

But is still don't know what to do after...

You need to differentiate the demand equation.  It is a bit more challenging here because of the fractional exponents.  You find the derivative either by using partial derivatives, as suggested by @callculus, or you can implicitly differentiate with respect to time.  I find that the latter approach gets a little confusing here, so, in this instance, I prefer the former.  The former approach allows you to break it into "smaller pieces", though the end result is the same.

You are also explicitly given information on how $x$ and $y$ are functions of time.  You can calculate the value of $x$ and $y$ at time = 5 using those functions, and you can calculate the values of $dx \over dt$ and $dy \over dt$ by taking the derivative of those functions.

Implicit Differentiation with respect to time

To implicitly differentiate the demand equation with respect to time, you first differentiate each variable with respect to itself, and then multiply by the derivative of that variable with respect to time.  This is the chain rule.

Because the the demand equation consists of the sum of two smaller expressions, the derivative sum rule says that we can simply add the derivatives of each expression.  That is,

$$\frac {d (u + v)}{dx} = \frac {du}{dx} + \frac {dv}{dx}$$

So, let's first differentiate $21000 - \frac {\sqrt{x}}{2}$ with respect to $x$.  You can rewrite that as $21000 - \frac {1}{2}x^{1/2}$.  Differentiating that gives you $- \frac {1}{4 \sqrt{x}}$.  We are differentiating with respect to time, therefore we need to multiply by the derivative of the inner function, $\frac {dx}{dt}$.

Now let's differentiate the second expression, $-11(0.3y-10)^{3/2}$, with respect to $y$.  If I have not made an error, the derivative is $- \frac {99}{20}*\sqrt{0.3y-10}$.  Again, we are differentiating with respect to time, therefore we need to multiply by the derivative of the inner function, $\frac {dy}{dt}$.

Put it all together for the derivative of the demand equation:

$$D(x,y) = 21000 - \frac {\sqrt{x}}{2} - 11(0.3y-10)^{3/2}$$
$$\frac {dD}{dt} = (- \frac {1}{4 \sqrt{x}} \frac {dx}{dt}) - (\frac {99}{20}*\sqrt{0.3y-10}* \frac {dy}{dt})$$

We have one equation and 5 unknowns, $\frac {dD}{dt}, x, y, \frac {dx}{dt} ~and~ \frac {dy}{dt} $.  We are attempting to solve for $\frac {dD}{dt}$, so we just need to figure out values for $x, y, \frac {dx}{dt} ~and~ \frac {dy}{dt}$.  As noted above, we have the formulas for $x$ and $y$. Therefore we can solve for the values of $x$ and $y$ at $t=5$, and we can differentiate those equations to find $\frac {dx}{dt}$ and $\frac {dy}{dt}$.

The values of $x$ and $y$ at $t=5$ are

$$x = 55200 + 100t$$
$$x = 55200 + 100(5)$$
$$x = 55700$$

$$y = 136+10 \sqrt{t}$$
$$y = 136+10 \sqrt{5}$$
$$y \approx 158.36$$

The derivatives of $x$ and $y$ with respect to time are:

$$\frac {dx}{dt} = 100$$
$$\frac {dy}{dt} = \frac {5}{\sqrt{t}}$$

Substitute known values in to the equation:
$$\frac {dD}{dt} = - \frac {1}{4 \sqrt{x}} *100 - \frac {99}{20}*\sqrt{0.3y-10}*\frac {dy}{dt}$$

$$\frac {dD}{dt} = - \frac {1}{4 \sqrt{55700}} *100 - \frac {99}{20}*\sqrt{0.3(158.36)-10}*\frac {5}{\sqrt{5}}$$

If I've made no errors, the result is

$$\frac {dD}{dt} \approx -67.89$$

Partial Derivatives

When taking partial derivatives, you explicitly break down the process into smaller steps, but it is exactly the procedure outlined above.  You take the derivative with respect to the variable itself (for example $x$) and you treat the other variable (in this case $y$) as constant.  So, when calculating the partial derivative with respect to $x$ the equation effectively becomes
$$D(x,y) = 21000 - \frac {\sqrt{x}}{2} - C$$
$$\frac {\partial D}{\partial x} = (- \frac {1}{4 \sqrt{x}}) - 0$$

And, similarly, when calculating the partial derivative with respect to $y$, the equation effectively becomes
$$D(x,y) = C - 11(0.3y-10)^{3/2}$$
$$\frac {\partial D}{\partial y} = 0 - (\frac {99}{20}*\sqrt{0.3y-10})$$

Note that the slightly different notation $\partial$ is used to signify a partial derivative.

Once the partials are calculated, you use the chain rule formula as suggested by @callculus to calculate the derivative of demand with respect to time, that is:

$$ \frac {dD}{dt} = \frac {\partial D}{\partial x} \frac {dx}{dt} + \frac {\partial D}{\partial y} \frac {dy}{dt}$$

which, of course, is exactly the same result that you would obtain with implicit differentiation.

Hint: The change of $D$ at $t$ is

$$\frac{dD}{dt}=\frac{\partial D}{\partial x}\cdot \frac{d x}{dt}+\frac{\partial D}{\partial y}\cdot \frac{dy}{dt}$$

In case of $\frac{dD}{dx}$ you differentiate $D(x,y)$ w.r.t $x$. After that you calculate the value of $x(t)$ at $t=5$. And then you insert that value into $\frac{dD}{dx}$.


How to calculate the area of a graph when $x$ and $y$ values are given? I have $x$ and $y$ values and I need to calculate the area of the graph with those $x$ and $y$ values.

How can I calculate that?

Please help me regarding this!

Say you have a function $f(x)$, and a set of domain values $\{a= x_0,x_1,x_2,\ldots,x_n=b\}$, where $x_{i+1}>x_i$. The points $x_i$ partition the $x$-axis into a discrete set of subintervals $L_i = [x_i,x_{i+1}]$. You can approximate the area between the function and the $x$-axis for a given subinterval by the following formula (the trapezoid rule):

$$ A_i =  |L_i|\times \frac{f(x_i)+f(x_{i+1})}{2},$$

where $|L_i|=x_{i+1}-x_i$ is the length of the subinterval. The total  approximated area between $a$ and $b$ is just the sum of all the areas,

$$A=\displaystyle\sum_{i=0}^{n-1} A_i.$$

You can find an interactive demo of the trapezoid rule here. Loosely speaking, the greater the value of $n$ the better the approximation of the area.

This is the formula you are looking for I suppose:

$$\text{Area}=\int_{a}^{b}{f(x)\:dx},$$

Where $a$ and $b$ are your limits of integration and $f(x)$ is the function of the graph.


Probability of impossible event. There is question in my book: Probability of impossible event is?

After reading the question my instant answer was $0$ and that was the answer given.

But then i thought other way, question is probability of impossible event, so there are two outcomes possible or impossible (event can be certain or impossible).

Therefore probability of impossible outcome is $\frac{1}{2}$.

Can that also be answer?

Your first answer is right. Your second argument only works to compute the probabilities of equally likely outcomes. So it's perfectly good to figure out the chances of heads coming up from a coin flip. But if one outcome is specified to be impossible, then by definition you are not working with equally likely outcomes and so the computation breaks down.

It depends what you mean by "Probability of impossible event".  This is an ambiguous phrase which would be interpreted in several different ways:

1) The event is known ahead of time to be not possible, therefore by definition in mathematics, the probability is defined to be 0 which means it can never happen.  I suspect most people on this site will go for this answer.  An example would be drawing a joker from a standard (thus jokerless) $52$ card deck of playing cards.  It aint gunna happen.

2) It is not known ahead of time if an event is possible or impossible and you are asking what are the chances it may be impossible.  For example, flipping a coin so it lands on its edge.  Some people think this is impossible since they seem to always say $50$% chance heads and $50$% tails but that is wrong cuz it depends where it "lands".  I've seen a coin land on its side and stay there.  What if you flip it and it lands in soft sand for example?

3) A certain event may seem impossible now but in the future it may be possible.  For example, it was thought to be impossible that a dirt bike (motorcycle) could ever do a double backflip ramp to ramp in competition but it has been done fairly recently (2006 X-games by Travis Pastrana).  Actually to me it seems more like a double reverse flip (facing forwards but flipping backwards).  Possible outcomes for this event would be success without injury, success but with injury, failure with injury, death....

Is there a time constraint on your event?  If not, then it may just take a while for the seemingly impossible event to happen and thus confirm it is possible.

So my answer is depending on what definition/interpretation of "impossible" you use, the probability can differ although I wouldn't know how to compute any interpretation other than the interpretation of truly not possible.


Is $\cot x = \tan (π/2 - x) $ true for any angle $x$? Is $$\cot x = \tan \Big(\frac{π}{2} - x\Big)$$ true even when $x$ is not an acute angle ?

True for all $x$ for which $\cot(x)$ is defined since

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

It is an identity by definition.. yes, it is true for all arguments.


Can an undefined function be said to be not convex? Is there a formal argument as to whether or not an undefined function can be considered "not convex?"

I think this is practically analogous to saying that I don't have a banana and therefore it doesn't taste good, which in a certain sense makes sense, but only if you're allowed to qualify something that doesn't exist.

An example that came up:

If we have

$f:(0,\infty)\mapsto\mathbb{R}$

$g:(-\infty,0)\mapsto\mathbb{R}$

then can $g \circ f$ be considered "not convex?"

The definition of convexity with which I am familiar is as follows (if you mean some other notion of convexity, please edit your question accordingly):

  Let $X$ be an interval in $\mathbb{R}$ (or, more generally, let $X$ be a convex subset of $\mathbb{R}^n$).  Then we say that $f : X \to \mathbb{R}$ is convex if
  $$ f(tx_0 + (1-t)x_1) \le tf(x_0) + (1-t)f(x_1) $$
  for all $x_0,x_1 \in X$.

Essentially, this says that the line segment joining any two points in the graph of $f$ must be above the graph of $f$.  Note that this definition can be extended to more general domains—I'm not addressing that here.

Now, note that when you say that a function is "undefined," what you are actually saying is that the domain of that function is empty.  The emptyset is convex (vacuously), thus we can make sense of the definition of a convex function.  So suppose that $f : \emptyset \to \mathbb{R}$ (this function will have an empty range, but the doesn't stop us from making the codomain anything we like).  Since there are no $x_0, x_1\in\emptyset$, we make vacuously conclude that
$$ f(tx_0 + (1-t)x_1) \le t f(x_0) + (1-t)f(x_1) $$
for all possible $x_0, x_1 \in \emptyset$.  Therefore any function with an empty domain will be (vacuously) convex.

For your particular example, if $f : (0,\infty) \to \mathbb{R}$ and $g : \mathbb{R} \to (-\infty, 0)$, then the composition $f\circ g$ is a function with empty domain, i.e.
$$ f\circ g : \emptyset \to \mathbb{R}.$$
By the above argument, it is necessarily convex.

Of course, convexity is a topological/geometric idea; even if the function is not regular you can define the concept.

E.G.

$f:\mathbb{R} \to \mathbb{R}$ s.t.

$f(x)=x^2$ if $x\in \mathbb{Q}$

is not regular but it is Jensen-Convex.


To what number $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}$ converges This question is from Ponnusamy and Silvermann's complex variables with applications Ch- 12 , subsection Infinite products.

Question: prove that $\prod_{n=2}^{\infty} \frac{n^3+1}{n^3-1}=2/3$.

I factored numerator and denominator to $\frac{(n-1) \times n^2+n+1 } { (n+1) \times n^2 -n+1}$ . But I think it's not possible to cancel denominantor and numerator in a way to get a pattren of cancelling due to factors $n^2-n+1$ and $n^2+n+1$ . So, I am struck on how should this be approached and unable to find any idea.
Kindly guide on how this should be approached.
Thanks!

Hint: The first five terms in the product $$\prod_{n = 2}^{\infty}\dfrac{n^3+1}{n^3-1} = \prod_{n = 2}^{\infty}\dfrac{(n+1)(n^2-n+1)}{(n-1)(n^2+n+1)}$$ are $\dfrac{3 \cdot 3}{1 \cdot 7}$, $\dfrac{4 \cdot 7}{2 \cdot 13}$, $\dfrac{5 \cdot 13}{3 \cdot 21}$, $\dfrac{6 \cdot 21}{4 \cdot 31}$, and $\dfrac{7 \cdot 31}{5 \cdot 43}$. When you multiply these out, a lot of terms cancel. In general, when you multiply the first $N$ terms, a lot of stuff cancels, and you can simplify what's left. Then take the limit as $N \to \infty$.


Limit of $\sin(3x)/x$ as $x$ goes to zero I was asked to help a student with this limit as X goes to zero.

$$\lim_{x \rightarrow 0}\frac{\sin \left(3x\right)}{x}$$

Note, I am able to solve it myself using L'Hopital's rule, just looking at a graph, or by the calculator method of sneaking up on the result by entering .1, .01, etc.

But, the student told me the teacher wanted him to use the 'algebraic' method. I am a Math Aide (i.e. an in house tutor) and often the students will not have a text, only notes from class and worksheets. So I have no reference, and no opportunity to check with the teacher. Any assistance for what method the teacher wants? Keep in mind, the students haven't learned derivatives yet.

The limit
$$\lim_{x \rightarrow 0} \frac{\sin(x)}{x} = 1$$
is an important and standard one, but proving it takes a bit of ingenuity. There is a beautiful answer here: How to prove that $\lim\limits_{x\to0}\frac{\sin x}x=1$?

Assuming that limit has been established, we can apply it to this problem. Note that
$$\frac{\sin(3x)}{x} = 3\frac{\sin(3x)}{3x}$$
If we let $u = 3x$ then notice that $u \rightarrow 0$ if and only if $x \rightarrow 0$, so
$$\lim_{x \rightarrow 0}\frac{\sin(3x)}{x} = 3\lim_{x \rightarrow 0}\frac{\sin(3x)}{3x} = 3 \lim_{u \rightarrow 0} \frac{\sin(u)}{u} = 3 \times 1 = 3$$

You could also write sin(3x) as sin(2x+x), which is equivalent to
sin(2x)cos(x)+cos(2x) sin(x).
Then you could fill in the double angles:
2sin(x)cos(x)cos(x)+[1-2sin^2(x)]sin(x)
Simplify to
2sin(x)[1-sin^2(x)]+sin(x)-2sin^3(x)
Which is
3sin(x) -4sin^3(x)
Now we have
lim_x->0[3sin(x) -4sin^3(x)]/x]
3lim_x->0[sin(x)] - 4lim_x->0[sin(x)/x]•lim_x->0[sin^2(x)]]
3(1)-4(1)(0)=3
Or you can simply use the Sandwich Theorem, though that is not considered an algebraic solution per say.


Open affine subscheme of affine scheme with different underlying set is non-isomorphic If I have an affine scheme $X=\operatorname{Spec}A$, and an open subscheme $Y=\operatorname{Spec}B$ which is also affine and whose topological space is different from $X$, does it follow that $X$ and $Y$ are non-isomorphic?
But if $X$ and $Y$ would be isomorphic, then we'd have $A\cong B$ and hence $Y$ and $X$ must actually be the same as subsets of $X$, because the prime ideals of $A$ are $1:1$ to the prime ideals in $B$. I think I'm confused here in this situation about the interplay of "being the same" and "being isomorphic".

Let $k$ be a non-trivial ring.
Consider $A = k [s_1, s_2, s_3, \ldots, t_1, t_2, t_3, \ldots] [s_1^{-1}, s_2^{-1}, s_3^{-1}, \ldots]$, that is, the polynomial ring over $k$ in countably infinitely many variables, with half of those variables made invertible by localisation.
Let $B = A [t_1^{-1}]$.
Then $A \cong B$ as $k$-algebras: there is a unique homomorphism $B \to A$ sending $t_1$ to $s_1$, $t_{n+1}$ to $t_n$, and $s_n$ to $s_{n+1}$, and it has an obvious inverse.
So $\operatorname{Spec} A \cong \operatorname{Spec} B$ as schemes over $\operatorname{Spec} k$.
But $A \ncong B$ as $A$-algebras, and $\operatorname{Spec} B$ as a scheme over $\operatorname{Spec} A$ is a non-surjective open immersion.
Moral: you have to pay attention to the category when asking whether two objects are isomorphic.

Question: "If I have an affine scheme X=SpecA, and an open subscheme Y=SpecB which is also affine and whose topological space is different from X, does it follow that X and Y are non-isomorphic?"
Answer: Two affine schemes $X:=Spec(A), Y:=Spec(B)$ are isomorphic iff $A\cong B$ are isomorphic rings. In particular this implies that $X$ and $Y$ have "isomorphic" topological spaces. Hence if $X \cong Y$ are "abstractly isomorphic", this implies the topological spaces are isomorphic.
In general two schemes (or ringed spaces) $(X, \mathcal{O}_X), (Y, \mathcal{O}_Y)$ are "isomorphic" iff there are maps $f:(X, \mathcal{O}_X) \rightarrow (Y, \mathcal{O}_Y), g:(Y, \mathcal{O}_Y)\rightarrow  (X, \mathcal{O}_X)$ with $f\circ g =Id,g \circ f=Id$. In particular the maps $f,g$ induce homeomorphisms of topological spaces $X \cong Y$. If a strict  open subscheme $U \subsetneq V$ is abstractly isomorphic to $V$ you end up with a topological space $U_t \subsetneq V_t$ realizable as a strict subspace of $V_t$, but where there is an abstract isomorphism $U_t \cong V_t$. This stuation is "paradoxical" - the topological space $V_t$ would then be homeomorphic to a strict subspace of itself.
Remark: "But if X and Y would be isomorphic, then we'd have A≅B and hence Y and X must actually be the same as subsets of X, because the prime ideals of A are 1:1 to the prime ideals in B. I think I'm confused here in this situation about the interplay of "being the same" and "being isomorphic"."
Answer: Here is an example (the affine line over an algebraically closed field) to show that there are no strict open subschemes $U \subsetneq \mathbb{A}^1$ with $U_t \cong \mathbb{A}^1_t$ a homeomorphism induced by a map of schemes. Hence in this situation you avoid such paradoxes.
Example: Let $k$ be an algebraically closed field and let $A:=k[x]$ be the polynomial ring in one variable $x$ over $k$. Let
$$f(x):=(x-u_1)^{l_1}\cdots (x-u_m)^{l_m}$$
with $u_i \neq u_j\in k$. Let $X:=Spec(A)$ and $U:=D(f) \subsetneq X$ be a strict open subscheme of $X$ and assume $X \cong D(f)$ is an abstract isomorphism of schemes over $k$, induced by a map of $k$-algebras $\tilde{\phi}: A \cong A_f$. The map $\tilde{\phi}$ induce a map of $k$-algebras
$$\phi:k[x] \rightarrow k[x]$$
with $\phi(x)=a(x)$ and $\phi(f(x)):=f(a(x))\in k^*$ a unit. It follows
$\phi(a(x)):=a_0\in k^*$ is a unit with $a_0\neq u_i$ for all $i$. Hence the map $\phi$ factors as follows
$$ k[x] \rightarrow k[x]/(x-a_0) \rightarrow k \subseteq k[x].$$
Hence the induced map $k[x]\rightarrow A_f \rightarrow A$ is the "constant map": The induced map of affine schemes
$$ D(f) \rightarrow X$$
identifies $D(f) \cong \{(x-a_0)\} \subseteq X$ as a topological space with one point which contradicts the assumption that $D(f) \subseteq X$ is an open subscheme.
There are problems related to infinite sets and maps of infinite sets - an infinite  set can be embedded as a strict subset of itself. In the above discussion we see that if we consider maps of schemes and open subschemes, in the case of the affine line we can not construct a strict open subscheme which is isomorphic to the affine line. Hence in this situation we avoid such "paradoxes". If you ignore the algebraic structure you can - as mentioned in the comments - easily construct such examples, but then you are not doing algebraic geometry, you are doing set theory.
For a discussion of this and the "Peano curve" see the following post:
"Infinity" in mathematics and an elementary question on dimension.
https://en.wikipedia.org/wiki/Dedekind-infinite_set


Using induction to prove that $2 \mid (n^2 − n)$ for $n\geq 1$ Use induction to prove that, for all $n \in \mathbb{Z}^+$, $2\mid (n^2 − n)$.

That is, I am supposed to use induction to prove that $(n^2 − n)$ can be divided by $2$ when $n$ is a positive integer.

I've tried the following:
$$\begin{split} (n+1)^2 − (n+1) &= (n+1)(n+1) - (n+1)\\
&=(n+1)(n-1)+2 - (n+1)\\
&=n^2 +n -n -1 +2 -n -1\\
&=n^2 -n,
\end{split}$$
but I'm not particularly good at maths so I have no idea if this is even correct, and if it is, what it even means.

Comment: As others have mentioned a number of times, induction is not at all necessary in this particular problem, but I am sure you hear that loud and clear by now. You probably just want to see how an induction proof would look. I've sketched out a proof below, but before you read it, I would encourage you to take a look at this post on how to write a clear induction proof. I imagine it would help you a good bit in both understanding how to write up your induction proofs clearly but also constructing your proofs. Following the instructions in that link will force you to understand your problem. That being said, see if you can follow the proof below (feel free to leave a comment if a point does not make sense).

For any positive integer $n$, let $S(n)$ denote the statement
$$
S(n) : 2\mid (n^2-n).
$$

Base step: For $n=1, S(1)$ gives $1^2-1 = 0$, and $2$ divides zero. Thus, $S(1)$ holds.

Inductive step: Let $k\geq 1$ be fixed, and suppose that $S(k)$ holds; in particular, let $\ell$ be an integer with $2\ell = k^2-k$. Then
\begin{align}
[(k+1)^2-(k+1)]&= (k^2+2k+1)-(k+1)\tag{expand}\\[0.5em]
&= (k^2-k)+2k\tag{rearrange/simplify}\\[0.5em]
&= 2\ell+2k\tag{by ind. hyp.}\\[0.5em]
&= 2(\ell+k)\tag{factor out $2$}\\[0.5em]
&= 2\eta\tag{$\eta=\ell+k. \eta\in\mathbb{Z}$}
\end{align}
This proves $S(k+1)$ and concludes the inductive step $S(k)\to S(k+1)$.

By mathematical induction, for each $n\geq 1$, the statement $S(n)$ is true. $\blacksquare$

About your attempt. You wrote:

  I've tried
  $$\begin{split} (n+1)^2 − (n+1) &= (n+1)(n+1) - (n+1)\\
&=(n+1)(n-1)+2 - (n+1)\\
&=n^2 +n -n -1 +2 -n -1\\
&=n^2 -n
\end{split}$$

This is not correct. Just plug in $n=1$ to see that $n^2-n=0$ and $(n+1)^2-(n+1)=2$, so these two expressions are not always equal.

But you can get
$$(n+1)^2-(n+1)=(n+1)[(n+1)-1]=(n+1)n=n^2+n=(n^2-n)+2n.$$
This means that the difference between the two expressions equals $2n$, therefore it is even and they have the same parity.

There is also an easy proof without induction. You just need to notice that $n^2-n=n(n-1)$ and one of the consecutive integers $n$ and $(n-1)$ is even.


Constant Presheaf not necessarily a sheaf. Proof? Let $\mathcal{X}$ be a topological space and $\mathcal{F}$ the constant presheaf, that assigns to each open set $\mathcal{U}$ of $\mathcal{X}$ the set $A$. The restriction map is the identity $A \rightarrow A$. I want to show that this presheaf is not necessarily a sheaf and I am trying to understand the argument in this wikipedia article: http://en.wikipedia.org/wiki/Constant_sheaf.
In particular, let $\emptyset = \cup_i \mathcal{U}_i, \, \mathcal{U}_i=\emptyset$, be a covering of the empty set. Take two sections over $\emptyset$, i.e. take $s,s' \in \mathcal{F}(\emptyset)=A$. The wikipedia article says that the restrictions of $s,s'$ to any $\mathcal{U}_i$ must be equal. I don't understand this. For example, since $\mathcal{U}_i \subset \emptyset$, then we have a restriction map $\mathcal{F}(\emptyset) \rightarrow \mathcal{F}(\mathcal{U}_i)$ which sends $s$ to $s$ and $s'$ to $s'$. So why $s=s'$?
Something else, possibly related to what I am missing: since the elements of $A$ are abstract, we can not really say anything about $s|_{\mathcal{U}_i}$, right?

Thanks.

The argument in the Wikipedia article states that you should take the empty covering of the empty set, i.e. your indexing set (where $i$ comes from) is the empty set. The empty union is the empty set, so it is a covering.

Then you pick any $s \neq s^\prime \in A$ and consider them to be sections over the empty set ($A$ needs to have at least two elements). The condition that they agree after restricting is an empty one (because there is no set to restrict to), so by the local identity axiom, you would need $s=s^\prime$, a contradiction.

I think you are taken the wrong definition of the constant presheaf. Since the presheaf $\mathcal{F}$ is a functor on the category of open sets of $X$, the constant presheaf means that $\mathcal{F}$ is a constant function, i.e., $\mathcal{F}(U)= \mathcal{F}(V)$ for all $U,V$ open in $X$.


How can I prove that the sum has value $1$? The answer of this question
Typicality of boundedness of entries of continued fraction representations
involves the infinite sum $$\sum_{k=1}^\infty -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}=1$$
Wolfram does not display $1$ if I enter the sum.
The sum converges very slowly :
I worked out
$$\sum_{k=n}^\infty -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}\approx\frac{1}{n\ln(2)}$$ for large $n$ as follows :
With the help of Wolfram I found out $-\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}\approx \frac{1}{k^2\ln(2)}$ for large $k$. So, to show the approximation, I only need to show $$\sum_{k=n}^\infty \frac{1}{k^2}\approx \frac{1}{n}$$ , which follows from the definite integrals squeezing the sum.

Is my argumentation correct ?
How can I show that the sum has value $1$ ?

I found an easier proof thanks to Graubner's answer based on the infinite product idendity

$$\prod_{k=1}^n\frac{(k+1)^2}{(k+1)^2-1}=\frac{2(n+1)}{n+2}$$ which can be easily proved by induction.

Hence, we get the partial sum

$$\sum_{k=1}^n -\frac{\ln(1-\frac{1}{(k+1)^2})}{\ln(2)}=\frac{\ln(\frac{2n+2}{n+2})}{\ln(2)}$$

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

  $\ds{\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} = 1:\ {\large ?}}$.

\begin{align}
&\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} =
-\,{1 \over \ln\pars{2}}\sum_{k = 1}^{\infty}
\ln\pars{k^{2} + 2k \over \bracks{k + 1}^{\,2}}
\\[5mm] = &\
-\,{1 \over \ln\pars{2}}\sum_{k = 1}^{\infty}
\bracks{-\int_{0}^{1}{\dd t \over \pars{k + 1}^{2} - t}} =
{1 \over \ln\pars{2}}\int_{0}^{1}\sum_{k = 0}^{\infty}
{1 \over \pars{k + 2 - \root{t}}\pars{k + 2 + \root{t}}}\,\dd t
\\[5mm] = &\
{1 \over \ln\pars{2}}\int_{0}^{1}
{\Psi\pars{2 - \root{t}} - \Psi\pars{2 + \root{t}} \over -2\root{t}}\dd t
\end{align}
$\ds{\Psi}$ is the
Digamma Function.
Note that $\ds{\Psi\pars{z} \equiv \totald{\ln\pars{\Gamma\pars{z}}}{z}}$. $\ds{\Gamma}$ is the
Gamma Function.

With $\ds{\root{t} \mapsto t}$
:
\begin{align}
&\sum_{k = 1}^{\infty}\braces{%
-\,{\ln\pars{1 - 1/\bracks{k + 1}^{\,2}} \over \ln\pars{2}}} =
-\,{1 \over \ln\pars{2}}\int_{0}^{1}
\bracks{\Psi\pars{2 - t} - \Psi\pars{2 + t}}\dd t
\\[5mm] = &\
-\,{1 \over \ln\pars{2}}\bracks{\vphantom{\Large A}%
-\ln\pars{\Gamma\pars{2 - t}} - \ln\pars{\Gamma\pars{2 + t}}}_{\ 0}^{\ 1} =
{1 \over \ln\pars{2}}
\ln\pars{\Gamma\pars{1}\Gamma\pars{3} \over \Gamma^{2}\pars{2}} = \bbx{\ds{1}}
\end{align}

  because $\ds{\Gamma\pars{1} = \Gamma\pars{2} = 1}$ and
  $\ds{\Gamma\pars{3} = 2! = \color{#f00}{2}}$.


what functions or classes of functions are $L^1$ but not $L^2$ We know that if a real function is in $L^2$ then it is in $L^1$, but the reverse is not necessarily true. So what are the examples of functions that are $L^1$ but not $L^2$, especially those intrinsically unbounded ones if any? Welcome more examples, or classes of examples. Thanks.

Neither containment holds in general.  On $\mathbb{R}$, the function $f$ such that $f(x)=1/x$ if $x\geq1$ and $f(x)=0$ otherwise is in $L^2\setminus L^1$, and the function $g$ such that $g(x)=1/\sqrt{x}$ if $0\lt x\leq1$ and $g(x)=0$ otherwise is in $L^1\setminus L^2$.

For bounded domains, $L^2\subset L^1$, because what keeps a function from being integrable on a bounded set is being too large, and squares make large numbers larger.  If $f$ is a bounded function in $L^1$, then $f$ is also in $L^2$, because what keeps a bounded function from being integrable is not going to zero fast enough, and squares make numbers go to zero faster.

Coming across these answers, it might be valuable to add the following explicit calculation:
Let $f \in L^1(\mathbb{R}^d) \cap L^{\infty}(\mathbb{R}^d)$. Then $f \in L^2(\mathbb{R}^d)$.
Indeed, denoting by $dx$ the Lebesgue measure on $\mathbb{R}^d$, we have
$$\int_{\mathbb{R}^d}|f|^2 = \int_{|f| \geq 1} |f|^2 + \int_{|f| < 1} |f|^2 \leq \int_{|f| \geq 1} C^2 + \int_{|f| < 1}|f| \leq C^2 \cdot dx(|f| \geq 1) + ||f||_{L^1} < +\infty.$$ The first inequality uses the essential boundedness of $f$ and the last inequality uses $f \in L^1(\mathbb{R}^d),$ which yields $dx(|f| \geq 1) < +\infty$.
On an unbounded domain $D$ (such as $D= \mathbb{R}^d$), the "converse", i.e. $L^{\infty}(D) \cap L^2(D) \subseteq L^1(D)$ does not hold: While we can estimate
$$\int_D|f| \leq \int_{D \cap |f| > 1} |f|^2+\int_{D \cap |f| \leq 1} |f| \leq ||f||_{L^2(D)}+dx(D\cap \{|f| \leq 1\}),$$we cannot conclude $f \in L^1(D)$, because the last summand is certainly not finite. As nicely explained in the above answers, this may be understood as insufficient fast decay to $0$ of $|f|$ on $D \cap \{|f| \leq 1\}$, while the decay of $|f|^2$ on the same set is fast enough.


Uncountable union of positive sets being negative Let $(X,\Sigma,\nu)$ be a signed measure space. A set $S \subseteq X$ is positive/negative  if for all subset $U \subseteq S, U \in \Sigma \rightarrow \nu(U) \geq 0$/$\nu(U) \leq 0$.

Is it true that an uncountable set of positive sets $(S_j)_{j \in M}$, where $|M| > |\mathbb{N}|$, may have a union that is negative?

I know that this is not possible when $M$ is countable, but is it possible to exhibit an example?

In search of closure, here is an answer.  Yes, as the original poster has defined the problem, an uncountable set of positive sets $(S_j)_{j∈M}$ may have a union that is negative.  That is because a positive set $S_j$ is defined by the original poster as one with $\nu(U)\ge 0$ for all $U\subset S_j$, and thus a set $S_j$ with $\nu(U)=0$ for all $U\subset S_j$ qualifies as positive (and also as negative).

As noted a few days ago, an example is where $[0,1]$ is endowed with the negative of Lebesgue measure, and the positive sets are $\{x\}$ for $x\in [0,1]$. Not only is $\nu([0,1])\le 0$, as required by the original poster's definition of negative, but actually $\nu([0,1])=-1$.  Likewise for many subsets $U\subset [0,1]$ we have $\nu(U)<0$, while for others we have $\nu(U)=0$, for example the singletons $\{x\}$ and also, say, $\mathbb{Q}\cap [0,1]$.

On the other hand, suppose we redefine positive to mean that $\nu(S_j)>0$ and $\nu(U)\ge 0$ for all subsets $U$, and we redefine negative to mean $\nu(S_j)<0$ and $\nu(U)\le 0$ for all subsets $U$.  It's pretty obvious that we can't have an uncountable union of positives that's negative as described, simply because $S_j\subset \bigcup S_k$ and so, if $\bigcup S_k$ is negative as defined, we must have $\nu(S_j)\le 0$.

The answer to the question
'Is it true that an uncountable set of positive sets $(S_j)_{j\in M}$  where $|M|>|N|$, may have a union that is strictly negative?''

is  no: Let $X=[-2,-1]\cup [1,2]$. Let $\nu_1$ be a counting measure in $[-2,-1]$ and let $\nu_2$ be - counting measure  in $[1,2]$.  We put $\sum=\{X +Y: X \in [-2,-1]~\&~Y [1,2]\}$. We put $\nu(X +Y)=\nu_1(X)+\nu_2(Y)$. The $\nu$ is signed measure. If $S_j$ is positive set then we have that $S_j \subset [-2,-1]$. Hence $\cup_{j \in M}S_j \subseteq [-2,-1]$. Since  $\cup_{j \in M}S_j$ does not contain any element of $[1,2]$ we  claim that it is not negative.


Roots of biquadratic equation This question also was a part of my today's maths olympiad paper:

If squares of the roots of $x^4 + bx^2 + cx + d = 0$ are $\alpha, \beta, \gamma, \delta$ then prove that:
$64\alpha\beta\gamma\delta - [4\Sigma \alpha\beta - (\Sigma \alpha)^2]^2 = 0$

I found value of $\alpha\beta\gamma\delta$ = $d^2$ and $\Sigma \alpha$ = -2b.
How to move on ?

Set $y=x^2$ and find an equation for $y$.

To start with $y^2+by+cx+d=0$ so that $c^2y=c^2x^2=(y^2+by+d)^2=y^4+2by^3+(2d+b^2)y^2+2dy+d^2$ Whence $$y^4+2by^3+(2d+b^2)y^2+(2d-c^2)y+d^2=0$$

The equation for $y$ has roots which are the squares of the roots of the original equation. You can use Vieta's relations with this.

Complete solution of the problem:

Given that α,β,γ & δ are squares of the roots of $x^4+bx^2+cx+d=0$ hence $\sqrt{α},\sqrt{β},\sqrt{γ}$  & $\sqrt{δ}$ are the roots of the given equation.
Thus, we have

sum of the roots
$$=\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ}=\frac{-B}{A}=\frac{0}{1}=0$$  &

sum of the products of two roots
$$= \sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ}=\frac{C}{A}=\frac{b}{1}=b$$
now, taking the square of the first expression, we get
 $$(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})^2=0 $$
 $$α+β+γ+δ+2(\sqrt{αβ}+\sqrt{αγ}+\sqrt{αδ}+\sqrt{βγ}+\sqrt{βδ}+\sqrt{γδ})=0$$

$$∑α+2b=0 $$  $∑α=-2b\tag1$

Now, the product of the roots
$$=\sqrt{α}  \sqrt{β} \sqrt{γ} \sqrt{δ}=\frac{E}{A}=\frac{d}{1}=d$$
$$ \sqrt{αβγδ}=d   $$ $$  αβγδ=d^2  \tag2$$

Now, satisfying the given equation $x^4+bx^2+cx+d=0$ by its roots $\sqrt{α},\sqrt{β},\sqrt{γ}$  & $\sqrt{δ}$  one by one, we get the following
$$\begin{align}
α^2+bα+c\sqrt{α}+d=0\\
β^2+bβ+c\sqrt{β}+d=0\\
γ^2+bγ+c\sqrt{γ}+d=0\\
δ^2+bδ+c\sqrt{δ}+d=0\\
\end{align}$$
Now by adding above four expressions column-wise, we get
$$α^2+β^2+γ^2+δ^2+b(α+β+γ+δ)+c(\sqrt{α}+ \sqrt{β}+\sqrt{γ}+ \sqrt{δ})+4d=0$$

$$α^2+β^2+γ^2+δ^2+b(-2b)+c(0)+4d=0$$
  $$α^2+β^2+γ^2+δ^2=2b^2-4d \tag3$$

Now by squaring both the sides of eq(1),we have
 $$(α+β+γ+δ)^2=(-2b)^2$$
$$α^2+β^2+γ^2+δ^2+2(αβ+αγ+αδ+βγ+βδ+γδ)=4b^2$$
By substituting the value from eq(3), we get
$$2b^2-4d+2∑αβ=4b^2 $$
$$2∑αβ= 2b^2+4d $$
$$  4∑αβ= 4b^2+8d \tag4$$

Now, by substituting all the corresponding values from eq(1), (2) & (4) in the LHS of the expression to be proved, we get

$$LHS=64αβγδ-[4∑αβ-(∑α)^2 ]^2=64d^2-[4b^2+8d-(-2b)^2 ]^2$$
$$ =64d^2-[8d]^2=64d^2-64d^2=0=RHS  $$


What exactly is a random disturbance in control theory? I'm embarrassed to even ask but...

I frequently see this word used in articles about dynamical systems, but not until now have I questioned what it really is. I understand that it is opposite of a deterministic system, but again, not very explanatory.

A random disturbance or noise in a dynamical system is a result of the error in the modelling of the system or the error due to sensor. For e.g. imagine that you are trying to model a complex dynamical system. But due to its complexity you cannot model it exactly. So, there is some error in your modelling. If you know what could be the possible error in your modelling then you call that error as random disturbance or noise. This type of error is called modelling error.

Similarly, when you are using a sensor to read some data, the sensor may give you some error. For e.g, if you are using a vernier calliper to measure something, then there may be error due to offset of zero. Similarly, if you are using an electro-mechanical sensor, then there may be some error in the reading of current or voltage. All such errors are called random disturbance or noise. This type of error is called observation error.


Asymptotics of $\int_0^{\infty} e^{-kx}(\sin(kx)+\cos(kx)) \frac{x^{4}}{\sinh(x^2)} dx$ at large $k$ While working on a physics problem I came across an integral of this form:
$$\int_0^{\infty} e^{-kx}(\sin(kx)+\cos(kx)) \frac{x^{4}}{\sinh(x^2)} dx$$

For my purposes the numerical evaluation of this integral is fine, and I have done that, but I started to wonder whether it is possible to derive the large $k$ asymptotics in a simple way. Numerically the asymptotics looks exponential, but can I get a simple formula somehow?

I tried a few things that I could think of, like integration by parts, but to no avail. Maybe some kind of saddle point approximation would work, but I guess you have to start by extending the integral to $\int_{-\infty}^{\infty}$, but I am not sure how. I don't have much experience with this kind of thing, can anyone help?

One approach, that does not work is this:

If we substitute $y=kx we get:

$$\int_0^\infty e^{-y} (\sin y + \cos y) \frac{y^4 / k^4}{\sinh(y^2/k^2) } dy/k = \frac{1}{k^3} \int_0^\infty e^{-y} (\sin y + \cos y) \frac{y^4/k^2}{\sinh (y^2/k^2)} dy$$

Now here, the function $e^{-y}$ makes a cut-off of the integral, so for large $k$ you might think that you can make a Taylor expansion in $\frac{y^4/k^2}{\sinh (y^2/k^2)}$, if you tried you would get as a leading term $c/k^3$ where

$ c = \int_0^\infty e^{-y} (\sin y + \cos y) y^2 dy$

But, this integral is zero, so this term that seems to be the leading asymptotic term is zero. If you try to use the higher order term in the Taylor expansion of  $\frac{y^4/k^2}{\sinh (y^2/k^2)}$ you will get contants that are:

$ c = \int_0^\infty e^{-y} (\sin y + \cos y) y^{2+4n} dy$

but all of these are zero. I guess this means that all of these power like terms are zero, so the integral goes to 0 faster than $O(1/k^n)$ for any $n$.

In this answer we try to find an asymptotic estimate for the integral

  $$
I(k)=\int_0^{\infty}e^{-kx}(\sin(kx)+\cos(kx))\frac{x^4}{\sinh(x^2)}
$$

as $k\rightarrow +\infty$

Simple trigonometric considerations bring the problem in the more conveniant form

$$
I(k)=\lim_{R \rightarrow\infty}I_R(k)=\lim_{R \rightarrow\infty}\sqrt{2}\Im\left[e^{i\frac{\pi}{4}}\int_0^{R}\underbrace{\frac{x^{4}}{\sinh(x^2)}e^{-kx+ikx}dx}_{f(x)}\right]
$$

The OP already showed that the boundary at $x=0$ will not contribute to the integral, so a more sophisticad approach is needed

To proceed we determine a path of steepest descent which is simply $x=te^{i\frac{\pi}{4}}$ in the first quadrant.

Deforming the integral into a triangle with the steepest descent path as hypotenuse (with small semicircles at $x=\sqrt{in}$ to avoid the singularities of $\sinh(x^2)^{-1}$) gives us according to Mr. Cauchy

$$
I_R(k)=\\I_{R e^{i\frac\pi4}}(k)+\sqrt{2}\Im\left[i\int_0^{R}f(R+iy)dy\right]+\sqrt{2} \Im\left[i\pi\sum_{R\geq n\geq 1}\text{Res}(f(x),x={\sqrt{i\pi n}})\right]
$$

here the first integral on the right is interpreted as a principal value integral. Therefore, taking limits $R\rightarrow \infty$ we observe $\lim_{R\rightarrow\infty}I_{Re^{i\frac\pi4}}(k)=0$ because $\Im(e^{i\pi/4}f(x))=0$ on this ray. Furthermore the contribution from the opposite leg of the triangle goes as $f(Re^{i\alpha})\sim 2 e^{-R^2}$ and can be safely dismissed.

$$
I(k)=\sqrt{2}\pi\Im\left[i\sum_{n\geq 1}\text{Res}(f(x),x={\sqrt{i\pi n}})\right]\sim\sqrt{2}\pi\Im\left[i\text{Res}(f(x),x={\sqrt{i\pi}})\right]
$$

Because residues with bigger $n$ are exponentially surpressed.

The remaining residue is easily calculated according to the residue formula for a simple pole.

Doing the algebra we end up with

  $$
I(k)\sim\frac{\pi^{5/2}}{\sqrt{2}}e^{-\sqrt{2\pi}k}+\mathcal{O}(e^{-\sqrt{4\pi}k})
$$

Path of integration

original path of integration (green), deformed path of integration (blue) and singularities (red) in the complex $x$-plane

Actually for large $k$ the integral seems to go to zero as we can't say anything about the $sin(kx)$ and $cos(kx)$ terms except that they are bounded so :
$$       |e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|\leq2e^{-kx}\frac{x^4}{sinh(x^2)}$$

Then you can show that $\frac{x^4}{sinh(x^2)}$ is bounded as it goes to zero continuously when $x\rightarrow\infty$.
$$       2e^{-kx}\frac{x^4}{sinh(x^2)}\leq Me^{-kx}$$
Finally you can split the integral between an indefinite integral and an integral on a compact :
$$\int_1^\infty|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx\leq M\int_1^\infty e^{-kx}dx=M\frac{e^{-ka}}{k}$$
The you have :
$$\int_1^\infty|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx +\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx\leq M\int_1^\infty e^{-kx}dx+\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx=M\frac{e^{-k}}{k}\int_0^1|e^{-kx}\frac{x^4}{sinh(x^2)}\left[sin(kx)+cos(kx)\right]|dx$$


Domain of $x^{\sin x+\cos x}$ Why domain of $x^{\sin x+ \cos x}$ is $x\ge 0$ while domain of $x^x$ is $x>0$
And if we take the natural logarithim for
 $f(x)= x^{\sin x+ \cos x}$
we get $\ln(f(x))=(\sin x+ \cos x) \ln(x)$ and if $x=0$  we will get  $\ln(0)$ ?

For $f(x)=x^x$, $\text{dom}(f)=\{x | x\in \mathbb{Z}_{-}\} \cup \{x| x \in \mathbb{R}_{+}\}$. But $0 \notin \text{dom}(f)$. In the case of $x^{\sin(x)+\cos(x)}$, you can calculate the value at $x=0$, because $0^x = 0$ $\forall x > 0$.

About the first expression: if you plug $x = 0$ in, you get $0^{0+1} = 0$. In the second you get $0^0$, which is indeterminate. That is why the second expression is not defined for $x=0$.

You cannot compose $\ln$ with $x^{\sin x + \cos x}$ since this expression at $0$ returns $0$. The logarithm is not defined at $0$.


use polar coordinates to evaluate $ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $ $$ \int_0^1 \int_0^{\sqrt{1-x^2}} e^{\sqrt{x^2 + y^2}} dydx $$ converting to polar coordinates :

$$ \int_0^1 \int_0^{\sqrt{1-r^2}} e^{\sqrt{r^2(cos^2 + sin^2)}} dydx $$
using trig substitution $cos^2 + sin^2 = 1$

$$ \int_0^1 \int_0^{\sqrt{1-r^2}} e^{\sqrt{r^2}} dydx = \int_0^1 \int_0^{\sqrt{1-r^2}} e^r drdx  = \int_0^1  e^{\sqrt{1-r^2} } - 1  \space drdx  = e^1 - 1 - e^1 = -1  $$ Is this logic correct ?

If you draw that region in the plane you'll see that its the part of the unit disk contained in the first quadrant i.e calling the domain $D$ then $D = \{(r, \theta): 0 \leq r \leq ?, 0 \leq \theta \leq ?\}$. Fill those in.
Now for the integrand, recall that you just change $x,y$ to their polar form and tac on the $r$ from the jacobian which accounts for the relationship between areas when making the coordinate change i.e,

$$\textbf{Area}= \int_{\theta=?}^{\theta=?} \int_{r=?}^{r=?} e^r \cdot r \ dr \ d\theta$$

What you have is a mess.

you can't say $\cos^2 + \sin^2 = 1$  you need some object.

When you convert to polar coordinates, your coordinate system will be in terms of $r, \theta.\ \ \ dx\ dy$ will be replaced

Start with.

$x = r \cos\theta\\y=r\sin\theta$

your limits.

$y \le \sqrt {1-x^2}\\
r\sin\theta \le \sqrt {1-r\cos\theta}\\
r^2\sin^2 \theta + r^2\cos^2 \theta le 1$
r le 1$

$y\ge 0\\
r\sin\theta \ge 0\\
\sin\theta \ge 0\\
0\le \theta \le \pi $

$x>0\\
\cos\theta \ge 0\\
0\le \theta \le \frac {\pi}{2} $

The jacobian...
I am not sure I am ready educate you on this.  Look it up.
$dy\ dx = r\ dr\ d\theta$

$\int_0^{\frac {\pi}{2}}\int_0^1 re^r \ dr\ d\theta$

$\frac {\pi}{2} (r-1) e^r|_0^1\\
\frac {\pi}{2}$


I have the pattern: 1 + 2 + 3 + 4 + 5 + 6, but I need the formula for it I'm writing some software that takes a group of users and compares each user with every other user in the group. I need to display the amount of comparisons needed for a countdown type feature.

For example, this group [1,2,3,4,5] would be analysed like this:

1-2, 1-3, 1-4, 1-5
2-3, 2-4, 2-5
3-4, 3-5
4-5

By creating little diagrams like this I've figured out the pattern which is as follows:

Users - Comparisons
2     -   1
3     -   3 (+2)
4     -   6 (+3)
5     -   10 (+4)
6     -   15 (+5)
7     -   21 (+6)
8     -   28 (+7)
9     -   36 (+8)

I need to be able to take any number of users, and calculate how many comparisons it will take to compare every user with every other user.

Can someone please tell me what the formula for this is?

You want to know how many ways there are to choose $2$ users from
a set of $n$ users.

Generally, the number of ways to choose $k$ elements from a set of
order $n$ (that is, all elements in the set are distinct) is denoted
by
$$
\binom{n}{k}
$$

and is equivalent to
$$
\frac{n!}{(n-k)!k!}
$$

In the case of $k=2$ the latter equals to
$$
\frac{n!}{(n-2)!2!}=\frac{n(n-1)}{2}
$$

which is also the sum of $1+2+...+n-1$.

For more information see Binomial coefficient and Arithmetic progression

This is kind of a pseudo code:

Say you have n number of people, and you labeled them.

for i in (1,2,3,...,n), person i need to compare with all the people who has a number larger (strictly), so person i need to compare (n-i) times.

so adding up would be
(n-1) + (n-2) + ... + 3 + 2 + 1...

which would be the sum from 1 to (n-1)


Question regarding galois correspondence in ${\mathbb R} /{\mathbb Q}$ ${\rm Aut }\left({\mathbb R}/{\mathbb Q}\right)$ is trivial ,  which implies there is no intermediate fields between field of rational and field of real numbers but we know there are infinitely many intermediate fields of $\mathbb R$ containing  $\mathbb Q$.

Why I am getting an contradictions? Please explain where I went wrong .

As the other answers point out, the problem is that $\mathbb R/\mathbb Q$ is not a Galois extension. Their arguments are, however, incorrect: being finite is not necessary for an extension $L/K$ to be Galois, and there is a rich theory of infinite Galois extensions.

A Galois extension is an algebraic extension that is normal and separable. Separability is fine here, but the other two conditions do not hold:

$\mathbb R$ is not an algebraic extension of $\mathbb Q$. In particular, $\mathbb R$ contains transcendental extensions such as $\mathbb Q(\pi)$
Even if we replaced $\mathbb R$ with $\mathbb R\cap\overline{\mathbb Q}$, the extension would still not be normal. For example, it contains the real root of $X^3-2$ but none of the complex roots.

For these two reasons, one cannot apply Galois theory to the extension $\mathbb{R/Q}$.

Galois correspondance works only for Galois extensions, that is finite-dimensional (which fails here) separable normal ones.


How to solve complex equations of the form $(z+a)^n=(z+b)^m$? I know how to solve complex equations of the form $(z+a)^n=r$ but how does the solution change when we have an equation of the form $(z+a)^n=(z+b)^m$?

Suppose we may represent $z+a=re^{i\phi}$ and $z+b=se^{i\psi}$, where $r, s, \phi, \psi$ can be calculated.
Then, our equation becomes $$r^ne^{in\phi}=s^me^{im\psi}$$
Equating like terms (we must, since the equation must be consistent in modulus, and the $e$ part has modulus $1$), we see:
$$r^n=s^m\implies \frac mn=\frac{\ln r}{\ln s}$$
$$n\phi+2k\pi=m\psi \implies \frac mn=\frac\phi\psi+\frac{2k}{n\psi\pi}$$
for some $k\in \Bbb Z$
We can solve this system for
$$n=\frac{2k\ln[s]}{\pi\ln[r^{\psi}s^{-\phi}]}\implies m=\frac{2k\ln[r]}{\pi\ln[r^{\psi}s^{-\phi}]}$$ and find a solution $(m, n)$ for each $k$


What comes next in the sequence? This is a challenge problem that I am given in class, but I never figured it out. It's just for fun.

What comes next (after the third picture)? At first, it seems like the blue figure is just shifiting to the right. But the third picture messes everything up.

I've tried just about everything you can imagine (even writing the blue dots as "1"'s and looking at the binary representation of the rows and columns), and I can't get it.

These are the choices.

I would guess (a) or (b), but I don't have solid reasoning.

This is the correct answer. The key here is to realize that the pattern is moving right one by one. So your first column becomes your second and your second becomes your third.

When they reach the third column, both the yellow and blue dots shift one down and if they are already at the bottom, they go to the top. (They appear at the first column again with these transformations).


Proof of Equation by Well Ordering Principle I have an assignment question

  Prove by either the Well Ordering Principle or induction that for all nonnegative integers $n$: $$\sum_{k=0}^n k^3 = \left(\frac{n(n+1)}{2}\right)^2.$$

I am able to solve this question using basic Induction, but not able to figure out how to do it by using Well Ordering Principle.

Any Solutions or hints would be very helpful.
Thanks in Advance

Suppose that the statement is not true.

Then the set $A=\left\{ n\in\mathbb{N}\mid\sum_{i=0}^{n}i^{3}\neq\left(\frac{n\left(n+1\right)}{2}\right)^{2}\right\} $
is not empty.

Since $\mathbb{N}$ is well-ordered set $A$ has a minimal
element $m$.

That means that $\sum_{i=0}^{n}i^{3}=\left(\frac{n\left(n+1\right)}{2}\right)^{2}$
is true for $n<m$ and is not true for $n=m$.

From this you can deduce
a contradiction.

(Start with $\sum_{i=0}^{m-1}i^{3}=\left(\frac{\left(m-1\right)m}{2}\right)^{2}$
and prove on base of that $\sum_{i=0}^{m}i^{3}=\sum_{i=0}^{m-1}i^{3}+m^{3}=\left(\frac{m\left(m+1\right)}{2}\right)^{2}$)

The conclusion is then that $A=\emptyset$ wich is
exactly the statement to be proven.

For $n=1$ the left side is equal to $1$ and the right side is also equal to $1$ so the equality holds true.
Suppose that the equality $(1)$ holds true for some $n\geq1$ then for $n+1$ we obtain $$\sum_{i=1}^{n+1} i^3  =\sum_{i=1}^{n} i^3 +(n+1)^3 = \frac{n^2 (n+1)^2 }{4} +(n+1)^3 =(n+1)^2\left(\frac{1}{4} n^2 +n+1\right) =(n+1)^2 \left(\frac{1}{2} n +1\right)^2 =\left(\frac{(n+1)(n+2)}{2}\right)^2 .\fbox{}$$


abstract Algebra (group theory) Three coins are placed on a table; showing heads. Can you get all the coins to show tails, by turning over two coins at a time? Use Group Theory to prove your answer.

I know that the answer is no I cannot get all coins to show tails however I  have no idea how to put it in terms of group theory. Any help would be appreciated.

You insist on using group theory to hide the parity check, so here it is.

In group-theoretic terms the situation is as follows: we have the group $G={\bf Z}_2^3$, and the subgroup $\langle(1,1,0),(1,0,1),(0,1,1)\rangle=H\leq G$, and we ask if $(1,1,1)\in H$.

The answer is no: consider the homomorphism $f\colon G\to {\bf Z}_2$ defined by $f(a,b,c)=a+b+c$. Clearly $H$ is contained in the kernel, while $(1,1,1)$ isn't, so $(1,1,1)\notin H$.

Alternately, you could just compose the generators of $H$ in all possible ways to see that actually $H=\{(0,0,0),(1,1,0),(1,0,1),(0,1,1)\}$, though this method doesn't generalise to a larger number of coins quite so well.

Group theory has nothing to do with this, it's a simple parity check.

There are three possibilities at each step:

you turn two heads into two tails: number of heads drops by 2, number of tails goes up by 2.
you turn two tails into two heads: number of tails drops by 2, number of heads goes up by 2.
you turn a head into a tail and a tail into a head: number of heads and tails is unchanged.

Since the numbers change by leaps of 2, if you start with an odd number of tails, you still have an odd number of tails after any whatsoever sequence of changes as before.

Obviously this applies to any odd numbered set of coins you decide to start with.


Given $n$, find $2$ positive integers $a,b$ such that $a+b=n$ and $LCM(a,b)$ is as minimum as possible This is a problem in a previous math/coding contest.
I broke this down into $2$ cases: $n$ is even or $n$ is odd. When $n$ is even, that means that $a=b=n/2$, because, intuitively, if we take $a < n/2$ (WLOG), then $b > n/2$, and so $LCM(a,b) \ge b > n/2$.
When $n$ is odd, I couldn't figure out how to do it. The only way I came up with was to brute force all values of $a$ (again, WLOG) from $1$ up until $n/2$, but this is way too slow when $n$ is large.
Any help would be much appreciated.

The answer is $a=k, b=n-k$ for odd $n>1$  and the largest proper (i.e., $<n$) divisor $k$ of $n$ and $a=b=n/2$ for even $n$.
Proof. Suppose $n$ is odd, $k+(n-k)$ is the decomposition of $n$ with the smallest  $LCM(k, n-k)$. Case 1. If $k$ divides $n$ then it divides $n-k$ and the $LCM$  is $n-k$, so in this case it is better to take $k$ to be the largest proper divisor of $n$.
Case 2. If $k$ does not divide $n$, then it does not divide $n-k$, so the $LCM$ is $k(n-k)/gcd(n,k)$. Let $k/gcd(n,k)=m$. Let $gcd(n,k)=d$. Then $k=md$. Then the decomposition is $md+(n-2d)$ and the $LCM$ is $m(n-2d)=mn-2md$. To be minimal we should have $mn-2md<n-d$, so $(m-1)n<(2m-1)d$, $d>((m-1)/(2m-1))n$. But since $n$ is odd and divisible by $d$ it implies $n\ge 3d$, so $3(m-1)/(2m-1)<1$, $m<3$ so $m=2$, which easily leads to a contradiction. Hence the minimum LCM is attained in Case 1 when $k$ is the largest proper ($\ne n$) divisor of $n$. $\Box$

If $a+b = n$ then $\gcd(a,b)|n$.
For any $d|n$ let $M_d$ be the minimum possible value of $\operatorname{lcm}(a,b)$ where $a+b = n$ and $\gcd(a,b) =d$.
Then we need to find $\min M_d$ for all $d|n$.
Let $d|n$.
We can find $a+b =n$ so that $\gcd(a,b) =d$ by letting $a=kd; k < \frac nd$ and $b=(\frac nd -k)d$.  If $\gcd(j,k)\ne 1$ we can always add or subtract values from $j,k$ (turns out this isn't that important so we don't need to go into detail).
$\operatorname{lcm}(a,b) = \frac {ab}{\gcd(a,b)} = jkd$.  Now $jk$ have the condition that $j+k = \frac nd = c$ a constant.  We can minimize $jkd$ but minimizing $jk$ which we can do by setting $j$ and $k$ as far apart as possible (that's just AM.GM. in revers); i.e. by setting $j=1$ and $k = \frac nd -1$ (and as $\gcd(1, \frac nd -1) =1$ that is why I said we wouldn't need to go into details).
Of for any given $d|n$ (and if $d=\gcd(a,b)$ then $d|n$) we can find a minimum such $ab$ so that $\operatorname{lcm}(a,b)= 1*(\frac nd -1)*d = n-d$.
So $M_d = n-d$.
And the minimum value of $M_d$ is $n-d$ where $d$ is the largest proper factor of $n$.
(We can't have $d = n$ because then $a,b \ge n$ and $a+b \ne n$.  But we can have $d=1$ if $n$ is prime.)
So in your case if $n$ is even then the largest proper divisor is $\frac n2$ and we have if $a= \frac n2*1$ and $b= \frac n2*(2-1)$ we have minimized it.
But any other $n = \prod p_i^{k_i}$ where $p_i$ are prime and $p_1 < p_2 .....$ then
$a = p_1^{k_1-1}\prod_{i>1}p_i^{k_i}$ and $b =  p_1^{k_1-1}\prod_{i>1}p_i^{k_i}(p_1-1)$ then $\operatorname{lcm}(a,b) = n- p_1^{k_1-1}\prod_{i>1}p_i^{k_i}$


Prove $[G:A] = [G:B]\cdot [B:A]$ if $A\leqslant B\leqslant G$ If $G$ is a group and $A\leqslant B\leqslant G$ then prove
  $$[G:A] = [G:B]\cdot [B:A]$$

For finite groups the theorem is obivious and simple to prove. But for infinte groups it seems a lot harder.

I've tried the following:

Let $L_G(A)$ be the set of left cosets of $A$ in $G$, then we need to prove
$$
|L_G(A)| = |L_G(B)|\cdot |L_B(A)|
$$
I was thinking of a trying to find a bijection $\alpha$ between $L_G(A)$ and $L_G(B) \times L_B(A)$. It seems logical to choose something like this:
$$
\begin{align*}
\alpha: L_G(A) & \to L_G(B) \times L_B(A)\\
gA & \mapsto (gB, b_gA)
\end{align*}$$
But how can I choose this $b_g$?

I know that $L_B(A)$ partitions $B$, this means that for any $ga \in gB$ (since $A\leqslant B)$ then $\exists b_i$ such that $ga \in g(b_iA)$ and then using this $b_i$. But is this useful?

Each coset $xB$ of $B$ in $G$ is the disjoint union of $|B:A|$ cosets
of $A$ in $B$. In detail, is $B=\bigcup_{i\in I} y_i A$ is a disjoint
coset decomposition, so is $xB=\bigcup_{i\in I} xy_i A$. Therefore
$G$ is the disjoint union $\bigcup_{j\in J} x_j B$
of $|G:B|$ $B$-cosets, and so a disjoint union
$\bigcup_{j\in J}\bigcup_{i\in I} x_j y_i A$
of $|G:B||B:A|$
$A$-cosets.

we can use the isomorphism theorem  $G/ B/B/A \simeq G/A$


Calculating the length of the paper on a toilet paper roll Fun with Math time.

My mom gave me a roll of toilet paper to put it in the bathroom, and looking at it I immediately wondered about this: is it possible, through very simple math, to calculate (with small error) the total paper length of a toilet roll?

Writing down some math, I came to this study, which I share with you because there are some questions I have in mind, and because as someone rightly said: for every problem there always are at least 3 solutions.

I started by outlining the problem in a geometrical way, namely looking only at the essential: the roll from above, identifying the salient parameters:

Parameters

$r = $ radius of internal circle, namely the paper tube circle;

$R = $ radius of the whole paper roll;

$b = R - r = $ "partial" radius, namely the difference of two radii as stated.

First Point

I treated the whole problem in the discrete way. [See the end of this question for more details about what does it mean]

Calculation

In a discrete way, the problem asks for the total length of the rolled paper, so the easiest way is to treat the problem by thinking about the length as the sum of the whole circumferences starting by radius $r$ and ending with radius $R$.
But how many circumferences are there?

Here is one of the main points, and then I thought about introducing a new essential parameter, namely the thickness of a single sheet. Notice that it's important to have to do with measurable quantities.

Calling $h$ the thickness of a single sheet, and knowing $b$ we can give an estimate of how many sheets $N$ are rolled:

$$N = \frac{R - r}{h} = \frac{b}{h}$$

Having to compute a sum, the total length $L$ is then:

$$L = 2\pi r + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi R$$

or better:

$$L = 2\pi (r + 0h) + 2\pi (r + h) + 2\pi (r + 2h) + \cdots + 2\pi (r + Nh)$$

In which obviously $2\pi (r + 0h) = 2\pi r$ and $2\pi(r + Nh) = 2\pi R$.
Writing it as a sum (and calculating it) we get:

$$
\begin{align}
L = \sum_{k = 0}^N\ 2\pi(r + kh) & = 2\pi r + 2\pi R + \sum_{k = 1}^{N-1}\ 2\pi(r + kh)
\\\\
& = 2\pi r + 2\pi R + 2\pi \sum_{k = 1}^{N-1} r + 2\pi h \sum_{k = 1}^{N-1} k
\\\\
& = 2\pi r + 2\pi R + 2\pi r(N-1) + 2\pi h\left(\frac{1}{2}N(N-1)\right)
\\\\
& = 2\pi r N + 2\pi R + \pi hN^2 - \pi h N
\end{align}
$$

Using now: $N = \frac{b}{h}$; $R = b - a$ and $a = R - b$ (because $R$ is easily measurable), we arrive after little algebra to

$$\boxed{L = 4\pi b + 2\pi R\left(\frac{b}{h} - 1\right) - \pi b\left(1 + \frac{b}{h}\right)}$$

Small Example:

$h = 0.1$ mm; $R = 75$ mm; $b = 50$ mm thence $L = 157$ meters

which might fit.

Final Questions:

1) Could it be a good approximation?

2) What about the $\gamma$ factor? Namely the paper compression factor?

3) Could exist a similar calculation via integration over a spiral path? Because actually it's what it is: a spiral.

Thank you so much for the time spent for this maybe tedious maybe boring maybe funny question!

The assumption that the layers are all cylindrical is a good first approximation.

The assumption that the layers form a logarithmic
spiral is not a good assumption at all, because it supposes that the
thickness of the paper at any point is proportional to its distance
from the center. This seems to me to be quite absurd.

An alternative assumption is that the layers form an Archimedean spiral.
This is slightly more realistic, since it says the paper has a uniform
thickness from beginning to end. But this assumption is not a much more
realistic than the assumption that all layers are cylindrical;
in fact, in some ways it is less realistic.

Here's how a sheet of thickness $h$ actually wraps around a cylinder.
First, we glue one side of the sheet (near the end of the sheet)
to the surface of the cylinder. Then we start rotating the cylinder.
As the cylinder rotates, it pulls the outstretched sheet around itself.
Near the end of the first full rotation of the cylinder, the
wrapping looks like this:

Notice that the sheet lies directly on the surface of the cylinder,
that is, this part of the wrapped sheet is cylindrical.

At some angle of rotation, the glued end of the sheet hits the part of
the sheet that is being wrapped. The point where the sheet is tangent to
the cylinder at that time is the last point of contact with the cylinder;
the sheet goes straight from that point to the point of contact with
the glued end, and then proceeds to wrap in a cylindrical shape around
the first layer of the wrapped sheet, like this:

As we continue rotating the cylinder, it takes up more and more layers
of the sheet, each layer consisting of a cylindrical section going
most of the way around the roll, followed by a flat section that joins
this layer to the next layer. We end up with something like this:

Notice that I cut the sheet just at the point where it was about to
enter another straight section. I claim (without proof) that this
produces a local maximum in the ratio of the length of the wrapped sheet
of paper to the greatest thickness of paper around the inner cylinder.
The next local maximum (I claim) will occur at the corresponding
point of the next wrap of the sheet.

The question now is what the thickness of each layer is.
The inner surface of the cylindrical portion of each layer of the
wrapped sheet has less area than the outer surface, but the portion of
the original (unwrapped) sheet that was wound onto the roll to make this layer had equal area on both sides. So either the inner surface was
somehow compressed, or the outer surface was stretched, or both.

I think the most realistic assumption is that both compression and stretching
occurred. In reality, I would guess that the inner surface is compressed more than the outer surface is stretched, but I do not know what the
most likely ratio of compression to stretching would be.
It is simpler to assume that the two effects are equal.
The length of the sheet used to make any part of one layer of the roll
is therefore equal to the length of the surface midway between the
inner and outer surfaces of that layer.
For example, to wrap the first layer halfway around the central cylinder
of radius $r$, we use a length $\pi\left(r + \frac h2\right)$
of the sheet of paper.

The reason this particularly simplifies our calculations is that the
length of paper used in any part of the roll is simply the area of the
cross-section of that part of the roll divided by the thickness of the paper.

The entire roll has inner radius $r$ and outer radius $R = r + nh$,
where $n$ is the maximum number of layers at any point
around the central cylinder. (In the figure, $n = 5$.)
The blue lines are sides of a right triangle whose vertices are
the center of the inner cylinder and the points where the first layer last touches the inner cylinder and first touches its own end.
This triangle has hypotenuse $r + h$ and one leg is $r$, so the other
leg (which is the length of the straight portion of the sheet)
is $$ \sqrt{(r + h)^2 - r^2} = \sqrt{(2r + h)h}.$$
Each straight portion of each layer is connected to the next layer
of paper by wrapping around either the point of contact with the glued
end of the sheet (the first time) or around the shape made by
wrapping the previous layer around this part of the layer below;
this forms a segment of a cylinder between the red lines with center at
the point of contact with the glued end.
The angle between the red lines is the same as the angle of the blue
triangle at the center of the cylinder, namely
$$ \alpha = \arccos \frac{r}{r+h}.$$

Now let's add up all parts of the roll. We have an almost-complete
hollow cylinder with inner radius $r$ and outer radius $R$,
missing only a segment of angle $\alpha$. The cross-sectional area of this is
$$ A_1 = \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2).$$
We have a rectangular prism whose cross-sectional area is the product
of two of its sides,
$$ A_2 = (R - r - h) \sqrt{(2r + h)h}.$$
Finally, we have a segment of a cylinder of radius $R - r - h$
(between the red lines) whose cross-sectional area is
$$ A_3 = \frac{\alpha}{2} (R - r - h)^2.$$
Adding this up and dividing by $h$, the total length of the sheet
comes to
\begin{align}
 L &= \frac1h (A_1+A_2+A_3)\\
   &= \frac1h \left(\pi - \frac{\alpha}{2} \right) (R^2 - r^2)
     + \frac1h (R - r - h) \sqrt{(2r + h)h}
     + \frac{\alpha}{2h} (R - r - h)^2.
\end{align}

For $n$ layers on a roll, using the formula $R = r + nh$,
we have $R - r = nh$, $R + r = 2r + nh$,
$R^2 - r^2 = (R+r)(R-r) = (2r + nh)nh$,
and $R - r - h = (n - 1)h$.
The length then is
\begin{align}
 L &= \left(\pi - \frac{\alpha}{2} \right) (2r + nh)n
     + (n - 1) \sqrt{(2r + h)h}
     + \frac{\alpha h}{2} (n - 1)^2\\
  &= 2n\pi r + n^2\pi h
      + (n-1) \sqrt{(2r + h)h}
      - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}\\
  &= n (R + r) \pi
      + (n-1) \sqrt{(2r + h)h}
      - \left( n(r + h) - \frac h2 \right) \arccos \frac{r}{r+h}.
\end{align}

One notable difference between this estimate and some others
(including the original) is that I assume there can be at most
$(R-r)/h$ layers of paper over any part of the central cylinder,
not $1 + (R-r)/h$ layers.
The total length is the number of layers times $2\pi$ times the
average radius, $(R + r)/2$, adjusted by the amount that is missing in the
section of the roll that is only $n - 1$ sheets thick.

Things are not too much worse if we assume a different but uniform ratio
of inner-compression to outer-stretching, provided that we keep the
same paper thickness regardless of curvature; we just have to make an
adjustment to the inner and outer radii of any cylindrical segment of the roll, which I think I'll leave as "an exercise for the reader."
But this involves a change in volume of the sheet of paper.
If we also keep the volume constant, we find that the sheet gets thicker
or thinner depending on the ratio of stretch to compression and
the curvature of the sheet.
With constant volume, the length of paper in the main part of the
roll (everywhere where we get the the full number of layers) is the
same as in the estimate above, but the total length of the parts of the
sheet that connect one layer to the next might change slightly.

Update: Per request, here are the results of applying the formula
above to the input values given as an example in the question:
$h=0.1$, $R=75$, and $r=25$ (inferred from $R-r=b=50$), all measured
in millimeters.

Since $n = (R-r)/h$, we have $n = 500$.
For a first approximation of the total length of paper,
let's consider just the first term of the formula. This gives us
$$
L_1 = n (R + r) \pi = 500 \cdot 100 \pi \approx 157079.63267949,
$$
or about $157$ meters, the same as in the example in the question.
The remaining two terms yield
\begin{align}
L - L_1
&= (n-1)\sqrt{(2r + h)h}
      - \left( n(r + h) - \frac h2 \right) \arccos\frac{r}{r+h} \\
&= 499\sqrt{50.1 \cdot 0.1} - (500(25.1) - 0.05)\arccos\frac{25}{25.1} \\
&\approx -3.72246774.
\end{align}
This is a very small correction, less than $2.4\times 10^{-5} L_1$.
In reality (as opposed to my idealized model
of constant-thickness constant-volume toilet paper), this
"correction" is surely insignificant compared to the uncertainties of
estimating the average thickness of the paper in each layer of a roll
(not to mention any non-uniformity
in how it is rolled by the manufacturing machinery).

We can also compare $\lvert L - L_1 \rvert$ to the amount of paper that
would be missing if the paper in the "flat" segment of the roll were
instead $n - 1$ layers following the curve of the rest of the paper.
The angle $\alpha$ is about $0.089294$ radians (about $5.1162$ degrees),
so if the missing layer were the innermost layer, its length would be
$25.05 \alpha \approx 2.24$, and if it were the outermost layer
it would be $74.95 \alpha \approx 6.69$ (in millimeters).

Just for amusement, I also tried expanding $L - L_1$ as a power
series around $h = 0$ (with a little help from Wolfram Alpha).
(To make $L - L_1$ a function of one variable $h$ with constants $R$ and $r$,
make the substitution $n = (R - r)/h$.)
This turns out to be a series of powers of $\sqrt h$ whose leading term is
$$
-\frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h.
$$
Plugging in the values from the example, this evaluates to
approximately $-3.7267799625$.
If you really wanted the length of the idealized toilet roll to the
nearest millimeter, but could tolerate an error of a few $\mu\mathrm m$
(for typical dimensions of a toilet roll),
a suitable approximation would be
$$
L \approx \frac{\pi (R^2 - r^2)}{h} - \frac{(R + 2r)\sqrt2}{3\sqrt r} \sqrt h.
$$

If there are $N$ layers, the thickness is $h=(R-r)/N$.

For cylindrical layers, the lengths increase linearly with layer number, so we can take the average circumference  $2\pi\bar r$, where $r= (r+R)/2$, times $N$:
$$
L = 2\pi \bar r N
$$

If we have an Archimedean spiral, the radius increases linearly with azimuthal angle. Thus, we can take the average radius $\bar r$ multiplied by the total angle $2\pi N$, and again:
$$
L = 2\pi N \bar r .
$$


Closed Form Solution for Minimization involving Standard Normal CDF and PDF Could someone please advice and provide detailed steps regarding any possible closed form solutions or other suggestions regarding solving a minimization problem of the type shown below? Here, $\phi$ and $\mathbf{\Phi}$ are the standard normal PDF and CDF, respectively. $\sigma$ is the standard deviation and $\theta$ is a parameter. We need to find the solution by expressing $x$ in terms of $y$ and the other parameters.

\begin{eqnarray*}
=\underset{\left\{ x\right\} }{\min}\left\{ x\sigma\left[\frac{\theta x}{\sigma}+\frac{\phi\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}\right]+\left(y-x\right)\sigma\left[\frac{\theta\left(y-x\right)}{\sigma}+\frac{\phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}\right]\right\}
\end{eqnarray*}

First Order Conditions (FOC) would give the below. Please point out ways to simplify or alternative approaches as well.

\begin{eqnarray*}
x\left[\frac{4\theta}{\sigma}+\theta\left\{ \frac{\phi'\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}-\phi'\left(\frac{\theta x}{\sigma}\right)\left[\frac{\phi\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}\right]^{2}\right\} \right.\\
+\left.\theta\left\{ \frac{\phi'\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}-\phi'\left(\frac{\theta\left(y-x\right)}{\sigma}\right)\left[\frac{\phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}\right]^{2}\right\} \right] & =\\
\left(\theta y\right)\left\{ \frac{2}{\sigma}+\frac{\phi'\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}-\phi'\left(\frac{\theta\left(y-x\right)}{\sigma}\right)\left[\frac{\phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}\right]^{2}\right\} \\
+\frac{\sigma\phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}{\Phi\left(\frac{\theta\left(y-x\right)}{\sigma}\right)}-\frac{\sigma\phi\left(\frac{\theta x}{\sigma}\right)}{\Phi\left(\frac{\theta x}{\sigma}\right)}
\end{eqnarray*}

Looks like a 1-variable problem.

For analytical approach , how about taking one derivative in $x$ and setting it equal to zero and see what you can get.

For a numerical approach, since you have the derivative (and even two if you like), you can call a minimization algo, or even rootfind the derivative with Newton's method or Brent or Secant method. This will, unfortunately, only work, for fixed values of $\theta, y$.


Algebraically solving the inequality $\frac{1}{x} - 1 > 0$ $$\frac{1}{x}-1>0$$

$$\therefore \frac{1}{x} > 1$$

$$\therefore 1 > x$$

However, as evident from the graph (as well as common sense), the right answer should be $1>x>0$. Typically, I wouldn't multiple the x on both sides as I don't know its sign, but as I was unable to factories the LHS, I did so. How can I get this result algebraically?

You have
$$\frac{1}{x} - 1 > 0$$
Forming a common denominator yields
$$\frac{1 - x}{x} > 0$$
The inequality is true if the numerator and denominator have the same sign.


$N$ be a natural number consisting all digits except $0.$ If the last digit of $N$ is 5 then prove that $N$ can't be a perfect square. Let $N$ be a nine digited natural number consisting all digits except $0.$ If the last digit of $N$ is 5 then prove that $N$ can't be a perfect square.

I have have tried finding a prime $p$ dividing $N$ but $p^2$ not dividing $N$ but I am unable to do it that way.  Also I found that ten's digit of $N$ can be $2$ or $7.$

Any help will be appreciated.

Tip: the last digit of $N$ is $5$, so $\sqrt N$ must also end with $5$, but if that's the case, the tens digit of $N$ must be $2$, since no perfect square ends with $75$, but if that's the case...

Let P be that number. After dividing by 5, u get Q. Now Q cannot contain 5. So Q doesn't have 5 as a factor. So Q times 5 cannot be a square as a square should have prime factors of order 2.

Edit:
Using the following facts which can be proved individually.

perfect square ending in 5, will also end in 25.
perfect square ending in 25 will have either 0,2,6 in hundred's place.
perfect square ending in 625, will have 0,5 in thousand's place.

so the number N, for it to be perfect square, should end in 625( it cant have 2 or 0 in hundred's place). After that it cant have eihter 0 or 5 in thousand's place. so N is not a perfect square.


Exercise 1.10 from Silverman "The Arithmetic of Elliptic Curves " I am having trouble with Silverman's exercise 1.10(b). The converse of (a) is easy because there is no integer solution to the equation when $p \equiv 3$ mod $4$. However, this method does not work for (b) because we always get empty set when intersecting whatever $V_p (p \equiv 3 $ mod $4)$ with $\mathbb{Q}$. I tried to intersect two varieties $V_p, V_q$ with something else, for example $\mathbb{Q}({\sqrt{p}})$ and try to tell the difference, but this is hard since it is very possible for $V_q$ to have nonempty intersection with $\mathbb{Q}({\sqrt{p}})$ and I think it is true that if so then $V_p, V_q$ are isomorphic over $\mathbb{Q}({\sqrt{p}})$. Are there some invariants I can use in this circumstance to show that two varieties are not isomorphic over $\mathbb{Q}$?

Using the Quadratic Reprocity Law, $\Big(\frac{p}{q}\Big)\Big(\frac{q}{p}\Big)=(-1)^\frac{(q-1)(p-1)}{4}=-1$, which implies that one of them (let's say p) is a square mod the other one, so $x^2+y^2=pz^2$ has solution in $\mathbf{F}_{q}$  because the equation $x^2+y^2=z^2$ has non trivial solution in $\mathbf{Z}$ but $x^2+y^2=qz^2$ has no solution in $\mathbf{F}_{q}$ because $q\equiv3\mod 4$ so they cannot be isomorphic over $\mathbf{Q}$.

You could calculate the Hilbert symbols $(\frac{p,-1}{q})_2$ for varying primes $q$.


Proving the existence of a non-decreasing sequence If $\{a_{n}\}$ is a sequence of positive numbers so that $\sum_{n=0}^{\infty} a_{n}$ converges. How can we show that there exists a non-decreasing sequence $\{b_{n}\}$ so that $\lim_{n\to \infty}b_{n}=\infty$ and $\sum_{n}^{\infty}a_{n}b_{n}$<$\infty$.

$\{b_n\}$ modifies the terms of the sequence. The trick is to modify chunks of the sum into terms of another convergent sequence. $\sum_{n=1}^{\infty}\frac{1}{n^2}$ is convergent, so we aim for this.

As $\sum_{i=0}^{\infty}a_i$ is convergent, there is a $k_n$ for all $n$ such that $\sum_{i=k_n}^{\infty}a_i<\frac{1}{(n)(n^2)}$. Then set $b_i=n$ for $i=k_n$ to $i=k_{n+1}-1$ for $n\geq1$ and $b_i=0$ otherwise.

Then $\sum_{i=k_n}^{k_{n+1}-1}a_ib_i<\frac{1}{n^2}$ for $n \geq 1$. Summing all these chunks gives a convergent sequence. And $b_n$ tends to $\infty$ as required.

$$ {a_n} = \frac{1}{ a^{3n }} $$

$${b_n} = a^{n} $$

for $ a>1 $, I think your requirements meet.

more explanation : in order to ensure $ \sum_{n}^{\infty} a_n$to converge,

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n} < p  $$ for some $ 0<p<1 $

find $ b$ that

$$ 1< \lim_{n\to\infty} \frac{b_{n+1}}{b_n} < \frac{1}{ p}  $$

so$ \sum_{n}^{\infty} b_n$ would diverge while nondecreasing, and

$$\lim_{n\to\infty} \frac{a_{n+1}}{a_n}\frac{b_{n+1}}{b_n} < \frac{p}{ p} = 1$$

which means $ \sum_{n}^{\infty} a_n b_n $ would converge


Does $f(n) = \Theta ( g(n ) )$ implies $f(n) \sim \gamma g(n)$ for $n \rightarrow \infty$ Saying that $f_n = \Theta(g_n)$ means there are two real constants $c_1,c_2$ and two integers $n_1,n_2$ such that

$$
c_1 g_n \leq f_n \leq c_2 g_n
$$

for all $n \geq n_0 = \max(n_1,n_2)$, I have the feeling that because of this there's a constant $c_1 < \gamma < c_2$ such that
$$
f_n \sim \gamma g_n.
$$

You can assume also that both $g(n)$ and $f(n)$ are strictly increasing sequences and that specifically

$$
c_1 g_n < f_n < c_2 g_n
$$

for $n \geq n_0$
My attempt to formalize was the following, for all $n$ I define the function

$$
c(\alpha)=c_1+(c_2-c_1)\alpha
$$
where $\alpha\in[0,1]$. And what I need to prove is that there's an $\alpha \in (0,1)$ such that

$$
f_n \sim c(\alpha)g_n
$$

as $n$ tends to $+\infty$. This implies I have to show

$$
\lim_{n\rightarrow+\infty} \frac{f_n}{c(\alpha)g_n}=1
$$

Or equivalently

$$
f_n = c(\alpha)g_n + o(g_n)
$$

for some $\alpha \in (0,1)$, but I don't actually know what to do... any clue? All I managed to prove is that $g_n$ can be multiplied by another sequence such that they're equivalent. But I can't prove that such sequence would converge.

No, it does not.

Counterexample 1: Here is a counterexample for functions:
$$
g(x)=3, \qquad f(x)=3+\sin x.
$$
and similarly for sequences:
$$
g_n=3, \qquad f_n=3+(-1)^n.
$$

Counterexample 2: Here is a more interesting counterexample for monotonic functions:
$$
g(x) = 3x, \qquad f(x)=3x+x\sin \log x, \qquad x\ge1;
$$
so
$$
f'(x) = (3 x + x \sin(\log x))' = \sin \log x + \cos \log x + 3 > 0.
$$

Similarly for monotonic sequences:
$$
g_n = 3n, \qquad f_n=3n+n\sin \log n, \qquad n\ge1.
$$

No, the claim is wrong because of the hidden constants.

Let $f(n)=2g(n)$. Then $f(n)=\Theta(g(n))$ but $f(n)\sim2g(n)\nsim g(n)$.


Why does the google calculator give $\tan 90^{\circ} = 1.6331779e^{+16}$? I typed in $\tan 90^{\circ}$ in Google and it gave $1.6331779\mathrm{E}16$. How did it come to this answer? Limits? Some magic?

Because it seems that Google Calculator works internally

in radians instead of degrees
with floating-point arithmetic instead of symbolic computation

The use of radians means that GC is using an angle of $\frac{\pi}{2}$ radians.  The nearest representable double value is $1.5707963267948965579989817342720925807952880859375$.  I shall denote this approximated value by $\frac{\pi}{2} - \epsilon$, where $\epsilon \approx 6.123233995736766 \times 10^{-17} $ (calculated by using a higher-precision value of $\pi$).

Recall that $\tan (\frac{\pi}{2} - \epsilon) = \cot \epsilon = \frac{1}{\tan \epsilon} $.  On my machine, this evaluates to $1.633123935319537 \times 10^{16}$.  (The  small-angle approximation of $\cot \epsilon \approx \frac{1}{\epsilon}$ happens to give the same answer.)  This is close to what Google Calculator returned, but differs by 33 ppm.


Why Special Limit of Euler's Number fails at higher powers? In the "Calculus 6th edition" by Edwards and Penney, in the chapter of Transcendental functions, there is an interesting question about special limit that leads to the famous Euler's number 2.718281828. It is given as:
$$\lim_{x\to\infty}(1+ \frac{1}x)^x$$
However, if you rise variable x to the higher power, say 10:
the graph literally goes crazy  as x goes to infinity, and the kills of to 1. Here is the graph of this situation:

This special limit states that: the further you go with x the closer you approach e. So, as you see the limit fails at higher powers. Please, help me to to understand this situation. I suppose it has something to do with the capability of computer systems to calculate.
Thank you!

You're experiencing floating point precision errors. They kill. The limit does not fail, but your computer does because it is using 64 bit floats behind the scenes, which have 53 digits available, which means they can only store numbers up to around $2^{53}\approx40^{10}$ so that's why your graph looks weird at that point. The weird behavior immediately before that point is caused by rounding to the nearest possible float, and when $x$ is too large $\frac{1}{x^{10}}$ gets rounded to $0$, so $\left(1+\frac{1}{x^{10}}\right)^{x^{10}}$ gets rounded to $1$.

This also happens at low numbers also. I tried with a power of 3 and it also happened just longer out in the x axis with a smaller jump


Recurrence relation in the bisection method When beginning to talk about error bounds on the bisection method for root finding, my book states the following:

  Let $a_n$ $b_n$ and $c_n$ denote the $n$th computed values of $a,b,$ and $c$, respectively. Then easily we get
  $$
b_{n+1}-a_{n+1}=\frac{1}{2}(b_{n}-a_{n}),\ n\ge1
$$
  and it is straightforward to deduce that
  $$
b_{n}-a_{n}=\frac{1}{2^{n-1}}(b-a),\ n\ge1
$$
  where $b-a$ denotes the length of the original interval with which we started.

However, wouldn't it be $b_{n}-a_{n}=\frac{1}{2^{n}}(b-a)$ instead? Simply take $n=1$. As stated, it wouldn't make sense that $b_{1}-a_{1}=(b-a)$, would it?

Informal statements on sequences are often a little ambiguous.

Sometimes a sequence is thought of as $a_0,a_1,a_2,\cdots$, sometimes as $a_1,a_2,a_3,\cdots$, and when you speak of the $n^{th}$ term, that can be $a_{n-1}$ or $a_n$. Here it is clear that numbering starts at $1$.

But another source of ambiguity is in "computed values". Shall we consider that the very first $a,b$ are computed, or are just given ? In the first situation, $a_1=a$. In the second, $a_1$ is a later value (as if $a=a_0$).

What we can say for sure: if we start with an interval of length $b-a$ before any iteration, every iteration halves that length and after $n$ iterations, it becomes $\dfrac{b-a}{2^n}$.


Trigonometric identity on $\cos \pi/7$ I found this in a book I used to train myself for the Math Olympics a bunch of years ago:

  Prove that $$\cos\frac{\pi}{7}-\cos\frac{2\pi}{7}+\cos\frac{3\pi}{7}=\frac{1}{2} $$

I couldn't solve it then and I can't solve it now.
As far as I know, polynomials might be involved somehow since it was in the polynomials section of the book, and it's not suposed to use complex numbers since it's in the section previous to that.

Using the double and triple angle formulaes I got the LHS to:

$$4\cos^3\frac{\pi}{7} -2\cos^2\frac{\pi}{7} - 2\cos\frac{\pi}{7}+1 = \frac{1}{2}$$

Using the sum-to-product formulaes I got to the same result.

And I'm pretty much stuck at this point.

There is a related question on the site with the answer found through this link: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=346908&sid=8ad587e18dd5fa9dd5456496a8daadfd#p346908

Answer I found useful was post #7

Older, yet same question:
How to prove $\cos\left(\pi\over7\right)-\cos\left({2\pi}\over7\right)+\cos\left({3\pi}\over7\right)=\cos\left({\pi}\over3 \right)$

The trick here lies in the '7'.   These are chords of a heptagon, and so if you mix $\cos \frac{4\pi}7 $ into the mix, you can use that these are equal, and thus derive the desired result.


What does in:variable1;in:variable2; mean? A question in my ODE textbook is as follows.

Determine whether the given first-order differential equation is linear in the indicated dependent variable.
$u dv+(v+uv-e^u)du = 0;$ in v; in u;

I feel as though I would be able to solve the question if I knew what the in v; in u; means.
In this question, what does in v; in u; mean? Is this a standard notation, and if so, what is it called?

This is really a question about English writing.  The question could be rewritten as

  Determine whether the first-order differential equation
  $$u dv+(v+uv-e^u)du = 0$$ is linear in $u$.  Determine also whether it is linear in $v$.

$u~dv+(v+uv-e^u)~du=0$

$u\dfrac{dv}{du}+(u+1)v-e^u=0$

$\dfrac{dv}{du}+\dfrac{u+1}{u}v=\dfrac{e^u}{u}$

$\therefore$ This is a first-order ODE with the dependent variable $v$ .


Can't understand a proof: Let $a,b,c$ be integers. If $a$ and $b$ divide $c$, then $lcm(a,b)$ also divides $c$ Can't understand a proof: Let $a,b,c$ be integers. If $a$ and $b$ divide $c$, then $lcm(a,b)$ also divides $c$

Proof I've been shown:
Since $c$ is a common multiple of $a,b$, it means that $lcm(a,b) \leq c$.

So:
$c=q\cdot lcm(a,b)+r$, $q \in \mathbb Z$ and $0\leq r \leq lcm(a,b)$

Since $c$ and $lcm(a,b)$ are common multiples of $a,b$, then so is $r$. But $m$ is the least common divisor, then it means $r=0$.

[Proof taken from: http://www.math.utah.edu/~wood/4400/hw1sol.pdf]

I cannot understand the last step... Why does it imply that $r=0$?

Thank you!

It follows from the (general) fact that if $a$ divides $b$ and $c$, then $a$ also divides $b-c$, so in this case:

$c = q.lcm(a,b) + r$ and $0 \leq r < lcm(a,b)$

$c - q.lcm(a,b) = r$

Which means that both $a$ and $b$ divides $r$, but $0 \leq r < lcm(a,b)$. So $r$ must be equal to $0$, and we get:

$c = q.lcm(a,b)$, as you wish.

I think it's easier using examples. if a=1, b=5, lcm=5.
If we pick c, a common multiple of 1 and 5, such as c=10 for example,
then c=q.lcm : in this case 10 = 2 * 5 + r
since lcm is the least common divisor, it means it is a direct multiple, so r=0 as seen from the example.


Difference between "property" and "structure" I sometimes hear that "$A$ is a property of $V$, not a structure."

But I am not sure how to distinguish "property" from "structure" in general. Could you explain the difference with some easy examples?

A property is something that is true of some object, it's intrinsic to the object. For example, being of finite dimension is a property of a vector space: either it's of finite dimension, or it's not. You can think of it as some kind of adjective, if you want: in "a blue pen", "blue" is a property of the pen.

A structure is something that you add on top of the object. To continue the same example, given a vector space $V$ of dimension $n$ (the dimension is a property of $V$), you can choose a basis $(e_1, \dots, e_n)$ of $V$, and consider the new object $(V, (e_1, \dots, e_n))$, a vector-space-with-a-basis. It's now a different kind of object.

If you have a structure $A$, not all objects can be upgraded to be an object-with-structure-$A$. For example, without the axiom of choice, some infinite dimensional vector spaces don't have a basis. So you have a property of objects, which is "being able to be given the structure $A$". For example, "having a basis" is a property of vector spaces, it's an intrinsic feature of the space. But the specific basis that you choose is not something intrinsic to the vector space, as there a vector space typically has many different bases, and each one is a different structure that you can add to you vector space.

Now, when you have a property $A$ that you can give to some type of objects, sometimes one can say "$A$ is a property, not a structure, of $V$". This means that $V$ can be given the structure $A$, and moreover it can be given that structure uniquely. So the structure that you can add to $V$ is actually intrinsic to $V$, there is only one possible given a $V$.

Here's an example: given a vector space, you can wonder whether or not you can make it into a topological vector space. It's not always possible in general, and if you can do it, there are several inequivalent ways of doing it. But when your space $V$ is of finite dimension, it's always possible to give a topology to $V$ that makes it into a topological vector space, and moreover that topology is unique (cf. this question). So what was a priori a structure on vector spaces becomes a property on finite dimensional vector spaces.

I'm not sure what context you are interested in but maybe this example helps:

Take the set $\mathbb{Z}$.  We can put a group structure on $\mathbb{Z}$ via the binary operation $\mathbb{Z} \times \mathbb{Z} \xrightarrow{+} \mathbb{Z}$ defined by $(n,m) \mapsto n+m$, so we could say $+$ gives a group structure on $\mathbb{Z}$.  A property of $(\mathbb{Z},+)$ would be something like 'the sum of two even numbers is even'.