Combinatorial proof of $a^n - b^n = (a - b)(a^{n – 1} + a^{n – 2}b + \dots + ab^{n – 2} + b^{n – 1})$ Is it possible to come up with a combinatorial argument which proves the following identity?

$$a^n - b^n = (a - b)(a^{n – 1} + a^{n – 2}b + \dots + ab^{n – 2} + b^{n – 1})$$

My idea was this:

Consider the number of ways $N$ to distribute $n$ numbered balls into $a$ numbered boxes such that there must be at least $1$ ball among the first $a - b$ boxes. The number of ways to do this is take the total number of ways without restriction, and subtract off the number of ways that all the balls are distributed only among the last $b$ boxes. That is,

$$N = a^n - b^n$$

Alternatively, we can first pick at least $1$ ball(s) and distribute them among the first $a - b$ boxes, and then distribute the remaining balls among the last $b$ boxes. That is,

$$N = \binom{n}{1}(a-b)b^{n-1} + \binom{n}{2}(a-b)^2b^{n-2} + \dots + \binom{n}{n}(a-b)^nb^0$$

However, this does not resemble the required $RHS$ of the identity.

Let ball $k$ be the highest numbered ball that gets placed in one of the first $a-b$ boxes.  Then all balls numbered $k+1$ through $n$ get placed in the last $b$ boxes, whereas balls numbered $1$ through $k-1$ may be placed in any of the $a$ boxes.

Using the binomial formula:

$a^n - b^n = (a-b+b)^n - b^n = \sum_{k=0}^{n-1}{n\choose k}(a-b)^{n-k}b^k = (a-b)(\sum_{k=0}^{n-1}{n\choose k}(a-b)^{n-k-1}b^k)$

And
$$\sum_{k=0}^{n-1}{n\choose k}(a-b)^{n-k-1}b^k = \sum_{k=0}^{n-1} \sum_{j=0}^{n-k-1}{n\choose k}{n-k-1\choose j}(-1)^ja^{n-k-1-j}b^{j+k} = \sum_{i=0}^{n-1}(\sum_{j+k=i} {n\choose k}{n-k-1\choose j}(-1)^j)a^{n-1-i}b^i$$

We need to prove $\sum_{j+k=i} {n\choose k}{n-k-1\choose j}(-1)^j = 1$ i.e. $ \sum_{j=0}^i{i\choose j}\frac{(-1)^j}{n-(i-j)} = \dfrac{1}{{n \choose i}(n-i)}$

This can be proven by partial fraction decomposition combined with the fact $\sum_{j=0}^n(-1)^{n-j}{n\choose j} = 0$.

The partial decomposition says
\begin{align}
\dfrac{1}{{n\choose i}} &= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-j}{n-j} \\
&=  -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-n}{n-j} - \sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{n-j}{n-j} \\
&= -\sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{i-n}{n-j} \\
\end{align}
which gives the conclusion
\begin{align}
\dfrac{1}{{n\choose i}(n-i)} = \sum_{j=0}^{i}(-1)^{i-j}{i \choose j}\frac{1}{n-j} = \sum_{j=0}^{i}(-1)^{j}{i \choose j}\frac{1}{n-(i-j)}
\end{align}

Clearly, if we do things in the inverse direction, we can prove the the partial fraction decomposition with $(a^n -b^n)= (a-b)(a^{n-1} + \cdots + b^{n-1})$.


Finding a continuous function and a sequentially compact Is there such a continuous function $f: R\rightarrow R $ and a sequentially compact $K \subset R$ such that the inverse $f^{-1}(K)$ is not sequentially compact?

Could someone provide me with some examples so I can get a better understanding of sequentially compact?

Hint: If you choose sin x as your function what would you choose for $K$? $[-\pi/2, \pi/2]$?


The sum of powers of two and two's complement – is there a deeper meaning behind this? Probably everyone has once come across the following "theorem" with corresponding "proof":
$$\sum_{n=0}^\infty 2^n = -1$$
Proof: $\sum_{n=0}^\infty q^n = 1/(1-q)$. Insert $q=2$ to get the result.

Of course the "proof" neglects the condition on $q$ for this formula, and the sum really diverges. However I now noticed an interesting fact:

If you use two's complement to represent negative numbers on computers, $-1$ is represented by all bits set. Also, sign extending to a larger number of bits (that is, getting the same number in two's complement representation on more bits) works by copying the left-most bit (also known as sign bit) into the additional bits on the left.

Now imagine that formally you sign-extend the number $-1$ to infinitely many bits. What you get is an infinite-to-the-left string of $1$s. Which, using the normal base-2 formula $n = \sum_k b_k 2^k$ (where $b_k$ is the bit k positions from the right, i.e. $b_0$ is the rightmost bit), that infinite string of $1$s translates into exactly the sum above! So in some sense we have an independent re-derivation of that equation.

Now my question is: Is there something deeper behind this? Somehow I cannot imagine it is just coincidential.

Yes. What you are doing is known as working in the $2$-adic numbers.

The $2$-adic numbers are equipped with a curious notion of distance given by the $2$-adic metric. In this metric, two numbers are close together if their difference is divisible by a large power of $2$. In particular, large powers of $2$ are very small. So relative to the $2$-adic metric the geometric series you wrote down really does converge, and the value it converges to really is $-1$.

Recursive Definition Not The Same Thing As The Limit

There is no paradox to be found.  The limit of a summation doesn't add enough terms to create a paradox.  In fact, to take the limit, you have to subtract the tail of the recursion off.  This is what will be demonstrated, that the value of some object $A$ can be a finite value, when the limit of a sequence that it expands to is not finite.

$$
Limit[A] = A - Tail[A]
$$

A sum of "all" powers of 2 is just a way to write -1; and you can never actually add the "last" term to cause the sum to become -1, so there is no contradiction.  It's basic recursion, as found all over computer science.  Focus on what a recursive definition actually says.  We are defining an object $A$ with this property.
$$
A = 1 + 2A
$$
$$
(A = 1 + 2A) - 2A
$$
$$
-A = 1
$$
$A$ is an object whose recursive definition says that it has an equal value to $-1$, though it may have other representations.  This is no more strange than $0.999... = 1.000$ due to the trailing 9 triggering a carry and giving them both the same value.
$$
A = -1
$$
Nothing particularly controversial happened there.  But -1 is special because the sum of powers of 2 is always 1 less than the next power of 2.  Just substitute equals for equals, and it stays consistent:
$$
A = (1 + 2(1 + 2(1 + 2(1 + 2A))))
$$
$$
A = 1 + 2 + 4 + 8 + 16A
$$
For every finite number of terms to expand, this recurrence is equal to $-1$ due to $A = -1$,
$$
A = (1 - 2) = (1 + 2 - 4) = (1 + 2 + 4 - 8) = \cdots
$$
So, very carefully, proceed without invoking a notion of infinity or twos complement computer representations.  These are true for any finite $N$,
$$
A = 1 + 2 + 4 + 8 + \cdots + (2^N)A
$$
$$
-1 = 1 + 2 + 4 + 8 + \cdots + (2^N)(-1)
$$
$$
-1 = 1 + 2 + 4 + 8 + \cdots + (-(2^N))
$$
Write a closed form for first $N$ in the recursive definition,
$$
A = (2^N - 1) + (2^N)A
$$
$$
A = (-1 + 2^N) + (2^N)A
$$
That last term added in is the tail.  $N$ can be as high or low as we like.
$$
Tail[A] = (2^N)A
$$
$A$, when defined recursively, solves for $-1$, regardless of what the limit of the sum is.  So define the limit of the sequence as actually subtracting that tail off.  Remember that the whole point of the integer $+\infty$ is that it isn't a number of iterations that you can perform.  So no sum is in danger of yielding a result of -1.  Remove the tail of the recursion off, because in a sum, we can add any number of terms, but there is no last term if you want the limit, rather than the value of $A$:
$$
Limit[A] = \lim_{N \to \infty} (A - Tail[A])
$$
Now here is the crucial part, where people that don't believe in recursion will argue with the result.  $A$ is just $-1$ by definition.  $A$ has a finite value.  But if we subtract $Tail[A]$, we are subtracting an arbitrarily large number.
$$
Limit[A] = \lim_{N \to \infty} (-1 - (2^N)(A)) = \lim_{N \to \infty}(-1 + (2^N)) = -1 + \lim_{N \to \infty} 2^N = \infty
$$

The really important idea here is that in general, the value of the recursively defined value is not necessarily the same as the limit of any indefinitely long sequence that you can expand it to.

$$
A \not= Limit[A]
$$

Part of the problem is that the use of "$\cdots$" and "$\sum$" are not as specific as a recursive definition.  There is no way to interpret recursion other than substituting equals for equals.  This is why many computer languages only give you recursion as the way to define iterative constructs; because the computer cannot make "common sense" judgments, or iron out ambiguity for you.

Ambiguity Of Sum Notation With $\infty$

Currently, if somebody writes this down, I'm not sure which interpretation they mean.
$$
\sum_{N}^{0 \cdots \infty}{2^N}
$$

If they say that $L \to \infty$, then the intent is confusing:
$$
A = \sum_{n}^{0 \cdots L}{2^n}
$$
$$
\sum_{n}^{0 \cdots L}{2^n} = 1 + \sum_{n}^{1 \cdots L}{2^n} = 1 + 2\sum_{n}^{0 \cdots L}{2^n}
$$
We said this, without meaning to say it:
$$
A = 1 + 2A
$$
If $L$ is finite, then that was false, because the summations expand to exactly $1 + L$ terms being added.  If $L$ is infinite, the intent was generally not to define it recursively as $A = 1 + 2A$.  But it's unsettling that we are not substituting true equals for equals (stronger than being equivalent values).  If we ensure that the summations add the exact same number of terms so that they are exactly equal definitions:
$$
\sum_{n}^{0 \cdots L+1}{2^n} = 1 + 2\sum_{n}^{0 \cdots L}{2^n}
$$
The above statement cannot be solved for $-1$ because they don't add the exact same number of terms (that can be as many as we want).  A computer is going to need to convert equal statements to statements that are exactly equal in every way before it can perform a substitution.  Being imprecise with $\infty$ makes this hard to do without human intervention.
$$
(\sum_{n}^{0 \cdots L}{2^n}) + 2^{L + 1} = 1 + 2\sum_{n}^{0 \cdots L}{2^n}
$$
$$
A_2 + 2^{L + 1} = 1 + 2 A_2
$$
This doesn't cause $A_2$ to solve for -1 now.  In fact, it subtracted the tail off of the recursion:
$$
A_2 = 1 + 2 A_2 - 2^{L+1}
$$
$$
-A_2 = 1 - 2^{L+1}
$$
$$
A_2 = -1 + 2^{L+1}
$$
If you try to sit down and write code to do these algebraic manipulations, problems like this are quite obvious; because the representation has to be converted to be exact before a substitution match can be made.  Even exact same statements with different side-conditions are not exactly equal for substitution purposes.  You must substitute equals for equals exactly (some equivalences are not as well defined enough to perform deterministicaly by a machine).

The correspondence with 2s complement isn't accidental.

Each bit represents an added in power of 2.  Even in a finite representation, "all ones in high bits" means the same thing as "no zeroes in high bits (to the left)"; meaning that it doesn't necessarily rely on a notion of infinity to work.

$$
A = -1 = 1 + 2 + 4 + 8 + \cdots = 1 + 2A
$$

This even works on the other side of the decimal point as well, where "all ones (to the right)" means the same thing as "no zeroes (to the right)" and triggers a ripple/carry just like in any finite machine representation.  That is why simply negating the bits negates the number if you also represent the low bits (due to the ripple/carry).

Negate every bit above and below the decimal point for the number -1, gives you this:
$$
A_3 = 1/2 + 1/4 + 1/8 + \cdots = 1
$$


Latin phrase for "accepting without proof" Is there a Latin phrase that would be used when accepting some statement without providing the proof of such a statement?

For example, say you are working on an elementary number theory proof, and you make the statement "since $p$ is odd, $p^2$ is odd, which we accept [Latin phrase for 'without showing proof']." Obviously this is a simple thing to prove, but in some cases it might be nice to acknowledge that a proof exists and we do not wish to show it.

Could ex facie be used in such a situation ("we accept ex facie that $q$ odd implies $q^2$ odd")? Or failing the existence of a Latin phrase, is there a way that sounds a little less crude than "without proof"?

It appears as though ex facie bears the intended meaning. Prima facie has a similar meaning.


Proving a removable discontinuity exists So I made up the following function as a thought exercise (for myself).

$$f(x) = 1\hspace{0.3cm} \forall\hspace{0.1cm} x = 1,2,3 \hspace{0.4cm}\text{and} \hspace{0.4cm} -1 \hspace{0.3cm}\forall \hspace{0.1cm}x = 4,5,6.$$

I have already proven that this function is not continuous.

But how would I go about proving (or disproving—honestly I have no idea) that the following limit exists:

$$\lim_{x\to 3}f(x)=  ?$$

With the usual (discrete) topology on the set $\{1,2,3,4,5,6\}$, the function is continuous (like every function with that domain), and the limit does not exist (since it is not even meaningful to ask about the limit at a point which is not an accumulation point of the domain).

For $x\to 3^-$, f(x) is not defined.
Similarly, For $x\to 3^+$, f(x) is not defined.
But, at $x=3$, $f(x)=1$

For limit to exist at $x=3$, function must attain a value (same value) at $x\to3^-$ and $x\to 3^+$, which may be different from value of function at 3.

For eg., consider $$\lim_{x\to3^-}=\lim_{x\to3^+}=p$$ And, $$f(3)=q$$
So, Limit exists at $x\to3$ and its value is p. $$\lim_{x\to3}=p$$

  This is called as missing point discontinuity, since, f(3) is not q, but p, i.e. a missing point.
   It comes under removable discontinuity.

edit:

Also, there are cases where a function is not defined for one side, in such cases, we go for one-sided limit.
For eg. Consider $$f(x)=\sqrt x$$ f(x) is not defined for $x\to 0^-$
So, $$\lim_{x\to 0}=f(0^+)=0$$ i.e., limit is equal to value of f(x) for $x\to 0^+$.

  This is a one-sided definition of limits, used when there is a domain restriction at boundary points for a function.


Integration of an infinite sum How do you calculate:  $$\int_0^1 \left(\sum_{n=0}^\infty x^{2 n + 1} t^n (1 - t)^n\right) dt \text{?}$$

We can write
$$I = \sum_{n=0}^\infty x^{2n+1}\int_0^1 t^n(1-t)^ndt$$
also by definition of beta function
$$B(a+1,b+1) = \int_0^1t^a(1-t)^bdt = \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+b+2)}$$
here $a=b=n$ and $n$ is an integer
this gives
$$I = \sum_{n=0}^\infty x^{2n+1}\frac{\Gamma(n+1)^2}{\Gamma(2n+2)}$$
$$I = \sum_{n=0}^\infty x^{2n+1}\frac{(n!)^2}{(2n+1)!}$$
By ratio test
$$\lim_{n\to\infty}\left| \frac{a_{n+1}}{a_n}\right|= r$$
it comes out that series converges for $x<2$.

The sum converges for $|x^2t(1-t)|<1$. After that, you are left with the integrand that looks like
$$
\frac{1}{1-x^2t+x^2t^2}
$$
Close the square to get
$$
\frac{1}{(t-0.5)^2 - 0.25+\frac{1}{x^2}}
$$
which is an integral of the form (after substituting $v=t-0.5$
$$
\int_{0}^{1}\frac{dv}{v^2-a^2}
$$


How to calculate this limit expression? I am having a difficult time with the limit:

$$\lim_{n\to \infty} \left(1+\log\left(\frac{n}{n-1}\right)\right)^n $$

How should I approach this problem?

Herein, we present a way forward that does not rely on differential calculus, but rather uses an elementary pair of inequalities and the squeeze theorem.  To that end we proceed.

PRIMER:
In THIS ANSWER, I showed using only the limit definition of the exponential function and Bernoulli's Inequality that the logarithm function satisfies the inequalities
$$\bbox[5px,border:2px solid #C0A000]{\frac{x-1}{x}\le \log(x)\le x-1}\tag1$$
for $x>0$.

Using $(1)$ with $x=\frac{n}{n-1}$, we find that
$$\frac{n+1}{n}\le 1+\log\left(\frac{n}{n-1}\right)\le \frac{n}{n-1} \tag 2$$
Finally, from $(2)$ we find that
$$\left(1+\frac1n\right)^n\le \left(1+\log\left(\frac{n}{n-1}\right)\right)^n\le \frac{1}{\left(1-\frac1n\right)^n} \tag 3$$
whereupon applying the squeeze theorem to $(3)$ yields the coveted limit
$$\bbox[5px,border:2px solid #C0A000]{\lim_{n\to  \infty} \left(1+\log\left(\frac{n}{n-1}\right)\right)^n=e}$$

By applying an exponential of a logarithm to the expression we get
$$\lim_{x\to\infty} \left( 1+\log \left( \frac{x}{x-1} \right)\right)^x=\lim_{x\to\infty} \mathrm{exp}\left( x\log \left( 1+\log \left( \frac{x}{x-1}\right)\right)\right)$$

We will show that $$L:=\lim_{x\to\infty} \left( x\log \left( 1 + \log \left( \frac{x}{x-1}\right) \right) \right)=1$$ and thus the result is $\lim_{x\to\infty} \mathrm{exp} (1)=\color{red} e$.

$$L= \lim_{x\to\infty} \left( x\log \left( 1 + \log \left( \frac{x}{x-1}\right) \right) \right)=\lim_{x\to\infty} \left(\dfrac{\log \left( 1+\log\left( \frac{x}{x-1}\right)\right)}{\dfrac 1x}\right)$$
so by de l'Hospital's rule
$$L=\lim_{x\to\infty} \left( \dfrac{-\dfrac{1}{x(x-1)\left( \log \left(\frac{x}{x-1} \right) + 1\right)}}{-\dfrac{1}{x^2}}\right)=\lim_{x\to\infty} \left( \dfrac{x}{(x-1)\left( \log \left( \frac{x}{x-1}\right) + 1\right)}\right)$$
Use de l'Hospital's rule again to get
$$L=\lim_{x\to\infty} \left( \dfrac{1}{\log \left( \frac{x}{x-1}\right) +1 - \frac 1x}\right)=\dfrac{\lim_{x\to\infty} 1}{\lim_{x\to\infty} \left( \log \left( \frac{x}{x-1}\right) +1 - \frac 1x\right)}=\frac 1 1=1$$
as desired.


Determine whether the following integral inequality holds Suppose that $\int_0^1 f dx = 0$. Is it true that
$$\left( \int_0^1 fg dx \right)^2 \leq \left( \int_0^1 g^2 dx - \left( \int_0^1 g dx \right)^2 \right) \left( \int_0^1 f^2 dx \right)$$
For context, this was stated as a true or false question on an old qualifying exam. I've tried noting that
$$ \int_0^1 fg dx  = \int_0^1 f (g-1) dx$$
Then, employing Cauchy-Schwarz gives
$$\left( \int_0^1 f g dx \right)^2 \leq \left( \int_0^1 g^2 dx - 2 \int_0^1 g dx +1 \right) \left( \int_0^1 f^2 dx \right)$$
From here, I thought I had it, since we have that $-2 \int_0^1 g dx + 1 \geq -\left( \int_0^1 g dx \right)^2$, but of course the inequality is going in the wrong direction. I tried to build some counterexamples, but everything I've tried has shown the inequality to be true. Any help is appreciated!

This is true. If $\int_0^1g(x)dx=0$, this is just Cauchy-Schwarz. Now note that neither side changes if you shift $g$ by a constant.

If probabilistic arguments are employed, we can simplify the arguments without writing integrals.

Suppose $X\sim U[0,1]$.  We rewrite the given inequality in terms of expected values:
$$E[f(X)g(X)]^2 \le var[g(X)]E[f(X)^2].$$
As @Jason points out, shifting $g$ by a constant doesn't affect the inequality (since $E[f(X)]=0$, so $E[f(X)(g(X)+k)]=E[f(X)g(X)+kf(X)]=E[f(X)g(X)]$), so the given inequality holds if and only if
$$E[f(X)(g(X)-k)]^2 \le var[g(X)-k]E[f(X)^2].$$
To apply the Cauchy-Schwartz inequality, we take $k=E[g(X)]$, so that $var[g(X)-k]=E[(g(X)-k)^2]$.  Then it's equivalent to
$$E[f(X)(g(X)-E[g(X)])]^2 \le E[(g(X)-E[g(X)])^2]E[f(X)^2].$$

This is the Cauchy-Schwartz inequality which holds for any random variables.


Solving a non-linear Partial differential equation $px^5-4q^3x^2+6x^2z-2=0$ I have to find out the complete integral of :
$$px^5-4q^3x^2+6x^2z-2=0$$
My attempt: Let $$f(x,y,z,p,q)=px^5-4q^3x^2+6x^2z-2$$
So, $$f_p=x^5,f_q=-12q^2x^2,f_x=5px^4-8q^3x+12xz,f_y=0,f_z=6x^2$$
Applying Charpit's method:
$$\frac{dx}{x^5}=\frac{dy}{-12q^2x^2}=\frac{dz}{px^5-12q^3x^2}=\frac{-dp}{5px^4-8q^3x+12xz+6px^2}=\frac{dq}{12x^2q^2} $$ $$\implies \frac{dy}{-12q^2x^2}=\frac{dq}{12x^2q^2}$$ $$\implies q=\sqrt{(y+c)}$$

where $c$ is an arbitrary constant.

I don't see here a way to find out $p$. Please help me solve this problem. Even a hint would be much help.

I'll appreciate any help towards it.Thanks in advance.

From given pde and by using the value q, we get
$P=\frac{4(y+c)^{\frac{3}{2}}}{x^3}-\frac{6z}{x^3}+\frac{2}{x^5}$
and subtitute the values $p$ and $q$ in $Z=pdx+qdy$ and use multiplier $e^{-\frac{3}{x^2}}$
and on integration we get complete required complete integral


Good Pre-Calculus book? I was reading this article and the author mentioned I should come here and get some advice. I'm 17, currently taking Pre-Calc in high schooling doing really good, but I feel like I'm not getting the most out of it. The teacher feels like he's more interested in covering chapters than getting us to understand things deeply and that worries me. The article says:

  Try to find a book where the author treats you as the intelligent, independent person you are, not as someone who has to take a course for a degree requirement...go to some math forums (like Math Overflow) and ask for book recommendations, telling them you want to become good at math and not just pass a required course; give them specific details and they can help find a book perfect for you.

So yeah asked on Math Overflow and was suggested to come here. I want to get better at math and really understand the concepts deeply and appreciate it like it was intended to. Any help I can get will be appreciated. Thanks!

The series of books Algebra, Functions and Graphs, Trigonometry, and The Method of Coordinates by I. M. Gelfand and various co-authors is an excellent way to supplement a pre-calculus course.  The books were written for advanced high school students taking correspondence courses with professors in the Soviet Union and are available in English translation.  The books are clearly written, supplement topics found in the typical pre-calculus text, and provide challenging problems.

Another good source is a series of Japanese books edited by Kunihiko Kodaira.  They include Mathematics I: Japanese Grade 10, Basic Analysis: Japanese Grade 11, and Algebra and Geometry:  Japanese Grade 11.  These books are also available in English translation.  The grade 10 book is for a required course roughly equivalent to pre-calculus.  Regular track students then take a course based on Mathematics II:  Japanese grade 11.  Mathematically inclined students take courses based on both the Algebra and Geometry and Basic Analysis texts.  The texts are a good source of challenging problems and contain material that will supplement what you would learn in a pre-calculus course.

There are many good textbooks for Calculus, such as

[1] Walter Rudin. Principles in Mathematical Analysis. Academic Press.

[2] Zorich. Mathematical Analysis, Vol. 1-2. Springer-Verlag.

[3] Григорий Михайлович Фихтенгольц. Course of Calculus. ?


Which groups with order 2ⁿ, n≤128 have small commuting probability and still high element order? I have found some formulae to calculate/estimate the commuting probability $P_G$ of some finite non Abelian groups in this thesis. For example: $S_n$, $D_n$, $Q_8$ among others. The options from the list seem to be either of prime/odd order, or with $P_G \approx \frac{1}{4}$.
That makes me wonder whether there is some power-of-two group that stands not far behind from the "anti-abelianess" of $S_n$.
And, I am also asking my self whether $S_n$ stablishes some kind of lower limit for $P_G$, even considering groups with order different from powers of two.
An example of power-of-two group with $P_G \ll \frac{1}{4}$ is the multiplication over unitriangular matrices (modulo).
However, it is still much higher than that of $S_n$ for similar set sizes.
I tried also direct product of dihedral groups and achieved even higher $P_G$ than with the matrices.
I have been putting some code and results here, in the case someone is interested in further investigation; see sections Abstract algebra module,
Commutativity degree of groups and Tendency of commutativity on Mn.
I am looking for a group that is algorithmically feasible to be implemented,
like $S_n$, $D_n$, $Z_n$ etc. which are straight forward to code and fast to calculate.
UPDATE:
It seems that groups of order $2^n$ are fated to be almost Abelian because prime power order implies nilpotent.
UPDATE 2:
My best attempt was a direct product of $D_{34}$ and $Z_{2^{128}-159}$ (I doubled the order to avoid half bits when importing 128 bits from legacy systems, among other benefits [and disadvantages]).
Unfortunately, another attempt was to use the matrices mentioned above in the place of S_{34}, but, in a small test with 45 bits the matrices already commuted with p = ~5e-06 after 40 million pairs.
The direct product of dihedral groups give formula values worse than the matrices.
The usefulness of all this is to combine identifiers that can be nicely operated within certain guarantees/properties.
PS.
Thanks to the help of a couple mathematicians, almost a year by now, I am almost not feeling in completely strange waters. I even learned what is a happy family - pun intended. [and monster group]

There exists arbitrarily large $2$-groups $G$ such that the center $Z(G)$ is cyclic of order two. A nice recursive construction of such is the wreath product tower $C_2$,
$C_2\wr C_2$, $(C_2\wr C_2)\wr C_2$, $\ldots$. IIRC your background is more on the computer science side, so it may be easier for you to get a handle of these as groups of graph automorphisms $G_n$ of the full binary tree $BT(n)$ of $n$ levels (so $2^n$ leaves). Like $n=4$:

It is easy to convince yourself of the fact that an automorphism of this graph must permute the $16$ leaves (the leaves are the only nodes of valence $1$). Also, an automorphism is fully determined if we know how it acts on the leaves. Furthermore, an automorphism must either intechange the two halves (two copies of a $3$-level binary tree), or be an automorphism of both half-trees. Consequently
$$
G_n=(G_{n-1}\times G_{n-1})\rtimes C_2
$$
with the extra factor $C_2$ being the cyclic group generated by "the halves changing automorphism" that we can specify to be the reflection w.r.t. the vertical axis of symmetry. This is what the above wreath product tower means.
Anyway, we see that

The orders of the groups $G_n$ satisfy the formula
$$|G_n|=2|G_{n-1}|^2.$$ Clearly $|G_1|=2$ (a cyclic group), so $|G_2|=8$
(the dihedral group $D_4$), $|G_3|=128$, $|G_4|=2^{15}$ et cetera.
For an element $g$ of $G_n$ to be in the center, it has to commute with the halves changing reflection as well as any tiny perturbation of one of the halves. The latter forces $g$ to be in the subgroup $G_{n-1}\times G_{n-1}$ that map the two half-trees to themselves. For its part, the former then forces $g$ to have the mirror image effect on both halves, so $g=(g',g')$. Then the latter requirement again forces $g'$ to be in the center of $G_{n-1}$. A simple piece of recursive thinking then forces $g$ to be the automorphism that swaps all the pairs of leaves sharing the same parent. The conclusion is that $Z(G_n)\simeq C_2$.
I asked about the non-abelianity of these groups from a different direction in this question, where I queried about the size of the group $G_n/[G_n,G_n]$. Knowledgable commenters told me that my hunch is correct
and $|G_n/[G_n,G_n]|=2^n$. This tells us that the largest abelian homomorphic image of $G_n$ has $2^n$ elements.

It seems that groups of order 2^n are fated to be almost Abelian because prime power order implies nilpotent.
A nilpotent group is a group that is "almost abelian"


What is a simple, physical situation where complex numbers emerge naturally? I'm trying to teach middle schoolers about the emergence of complex numbers and I want to motivate this organically. By this, I mean some sort of real world problem that people were trying to solve that led them to realize that we needed to extend the real numbers to the complex.

For instance, the Greeks were forced to recognize irrational numbers not for pure mathematical reasons, but because the length of the diagonal of a square with unit length really is irrational, and this is the kind of geometrical situation they were already dealing with. What similar situation would lead to complex numbers in terms that kids could appreciate?

I could just say, try to solve the equation $x^2 + 1 = 0$, but that's not something from the physical world. I could also give an abstract sort of answer, like that $\sqrt{-1}$ is just an object that we define to have certain properties that turn out to be consistent and important, but I think that won't be entirely satisfying to kids either.

I don't know

  a simple, physical situation where complex numbers emerge naturally

but I can suggest a way to help you

  teach middle schoolers about the emergence of complex numbers and I
  want to motivate this organically.

I did this once as a guest lecturer in a middle school classroom by developing a geometric  interpretation of arithmetic on the number line.

Adding a fixed number $r$ is a shift by $r$, to the right if $r > 0$, to the left if $r < 0$. Successive shifts add the shift amounts.  Each geometric shift is characterized by the position that $0$ moves to. You illustrate this visually by physically shifting a yardstick along a number line drawn on the board.

The answer to the question "what do you shift by so that doing it twice shifts by $r$?" is clearly $r/2$.

This is looking ahead to square roots, but you don't say that yet.
The underlying idea is that the group of shifts is the additive group of the real numbers, but you don't say that ever.

Now that addition is done you go on to multiplication. Multiplying by a fixed positive $r$ rescales the number line. If $r>1$ things stretch, if $r < 1$ they shrink and multiplying by $r=1$ changes nothing.  To know what a scaling does all you need to know is the image of $1$.

Successive scalings multiply, just as successive shifts add.
What should you do twice to scale by $9$? Half of $9$ doesn't work, but $3$ does. The class will quickly grasp that the geometric way to halve a scaling is to find the square root.

What about multiplication by a negative number? The geometry is clear: it's  reflection over $0$ followed by a scaling by the absolute value. Again the transformation is characterized by the image of $1$.

Now you're ready for the denoument. What geometric transformation can you do twice to move $1$ to $-1$ on the number line? Take your yardstick, place it on the line on the board, rotate by a quarter of a circle so that it's vertical, then another quarter and you're there. The image of $1$ is not on the line. It's at position $(0,1)$ in the cartesian coordinate system middle schoolers know about. They will find it cool to think of that point as a new number such that multiplying by it twice turns $r$ into $-r$. Name that number "$i$".

If you have brought the class along this far the rest is easy. They will quickly see the $y$ axis as the real multiples of $i$. Clearly adding $i$ should be a vertical translation by one unit. Vector addition for complex numbers follows quickly. Ask for the square root of $i$ and they will rotate the yardstick $45$ degrees. If they know about isosceles right triangles they will know that the (actually a) square root of $i$ is $(\sqrt{2}/2)(1+i)$, which they can check formally with the distributive law (which they will not ask you to prove).

A caveat. I think this should be pure fun for the class. Make that clear, so if some don't follow they don't worry. I would not try to integrate it into whatever the standard curriculum calls for. It should probably not extend over multiple class periods. Save it for a day near the end of the school year.

Not 100% sure if it is a good example, but afaik the time-dependent Schrödinger equation contains i (https://en.wikipedia.org/wiki/Schr%C3%B6dinger_equation#Time-dependent_equation).


Composition of Number Fields If we have number fields $K$ and $L$ containing $\mathbb{Q}$ such that $K\cap L = \mathbb{Q}$, then is it true that $[LK:\mathbb{Q}] = [L:\mathbb{Q}][K:\mathbb{Q}]$?

This fact is true if both $K$ and $L$ are Galois extensions of $\mathbb{Q}$, because then the embedding
$$
Gal(LK/\mathbb{Q}) \hookrightarrow Gal(K/\mathbb{Q})\times Gal(L/\mathbb{Q})
$$
is an isomorphism.
It is not true without the Galois hypothesis : Consider $K = \mathbb{Q}(\sqrt[3]{2})$ and $L= \mathbb{Q}(\omega\sqrt[3]{2})$, then $K\cap L = \mathbb{Q}$, but $[KL:\mathbb{Q}] = 6$, and $[K:\mathbb{Q}][L:\mathbb{Q}] = 3\cdot 3 = 9$.

Certainly not. For example, if $K=L$ and $[K:\mathbb{Q}]>1$, the statement is clearly false. More generally, if $K\subseteq L$, the statement is false. For example, if $K = \mathbb{Q}(\sqrt{2})$ and $L = \mathbb{Q}(2^{1/4})$, then $[LK:\mathbb{Q}] = [L:\mathbb{Q}] = 4$ and $[K:\mathbb{Q}] = 2$.

In general, $[LK:\mathbb{Q}] = [LK:K][K:\mathbb{Q}]$, so that your formula holds if and only if $[LK:K] = [L:\mathbb{Q}]$.


Is there a polynomial time algorithm determine whether a given point $x$ is a basic feasible solution(vertex solution) of polyhedron $P$ Given a polyhedron $P \subset \mathbb{R^n}$ defined by $m$ inequalites $Ax \leq b$. Given a point $z \in \mathbb{R^n}$, is there a polynomial time algorithm to determine if $z$ is a basic feasible solution(vertex solution) of $P$?

Yes, just compute $Az.$ If it is equal to $b,$ then check how many constraints are violated for $z + \epsilon A_i$ (where $A_i$ is the $i$-th row.)


[[complex]] Solutions of $x=\sqrt{\sin{x}}$ I wanted to make one of those cool infinite recursive definitions for myself, and I chose one that I thought looked cool: $x=\sqrt{\sin{x}}=\sqrt{\sin{\sqrt{\sin{\sqrt{\sin{...}}}}}}$ for no other reason than because I thought it looked cool. Using my method* of finding solutions for these; start with a number e.g. 1, take sqrt(sin(1)), take sqrt(sin(that)) alot of times using Answer button until you find a good guess, then algebraically confirm said guess; I got as far as $0.8767262154$ on my TI-83 Plus and $0.87672621539$ on the calculator you get on http://www.google.com/search?q=calc and also don't forget $0$ courtesy of Mac Grapher.app,

wolfram alpha roasted all with a massive $0.876726215395062445972118643142$

What my real question is, is how would I calculate (not just compute) or solve for all the real/complex solutions of this? Also, my complex question no pun intended is, wolframalpha also gave me $x=\frac{\sqrt{i\left(e^{-ix}-e^{ix}\right)}}{\sqrt{2}}$, how did they get this?[[What are the complex solutions of this equation?]]

*I discovered this for myself when testing that $φ=1+\frac{1}{φ}$ but am aware that it may have already occured to people who know about Mandelbrot set and higher degree 2-dimensional polynomial equations for which algebraic methods of calculation hav yet to be developed

For the complex looking thing,

Remember that
$$e^{ix}=\cos x+\sin x$$
$$e^{-ix}=\cos x - \sin x$$

So $$e^{-ix} -e^{ix}=-2\sin x $$

Hence

\begin{align}
x&=\sqrt{\sin x}\\
&=\sqrt{\frac{e^{-ix} -e^{-ix}}{-2}}\\
&=\frac{\sqrt{i(e^{-ix} -e^{ix})}}{\sqrt2}
\end{align}

So cmarangu as for your complex solutions there are ~3.

Theres 0i.

And then theres one around 0.5231i

And then theres one around 4.3069i

And don't forget ∞i.

I found these solutions using Mac Grapher app I said y=x and found intersections with y=sqrt(sin(ix)) if ya know what I mean.

In order to find the exact values for these you may have to do some weird stuff with e^(ix) and ya.


Is it possible for an antiderivative to exist but not to be equal to the integral? This question is about:

$$\int_0^{2\pi} \frac{1}{1.1+\cos x}$$

The solution is $20\pi/\sqrt{21}$.

But when using the antiderivative,given here
I get a solution of $0$.

What is going on here?

The basic problem is that although there exist continuous antiderivatives, the antiderivatives given by a closed form formula have discontinuities (in this case at points where the tangent function goes to $\pm \infty$).  You have to take those into account when using the Fundamental Theorem of Calculus.

EDIT:
In this case the antiderivative given is
$$ F(x) = \dfrac{20}{\sqrt{21}} \arctan\left( \dfrac{\tan(x/2)}{\sqrt{21}}\right)$$
The problem is that $\tan(x/2)$ has a singularity at $x = \pi$: $\tan(x/2)$ is not defined there.  If you plot it, you'll see that
$F(x)$ actually has a jump discontinuity:

As $x$ approaches $\pi$ from the left, $\tan(x/2)/\sqrt{21}$ goes to $+\infty$, so arctan of that goes to $+\pi/2$, and $F(x)$ goes to $10 \pi/\sqrt{21}$.
As $x$ approaches $\pi$ from the right, $\tan(x/2)/\sqrt{21}$ goes to $-\infty$, so arctan of that goes to $-\pi/2$, and $F(x)$ goes to $-10 \pi/\sqrt{21}$.  Of course, the discontinuity means $F(x)$ is not really an antiderivative at $x = \pi$: a discontinuous function can't be differentiable.
In order to get a correct antiderivative, you have remove the discontinuity, e.g.
by adding $20 \pi/\sqrt{21}$ to the function on the right of $x=\pi$.
Thus a correct antiderivative over this interval is
$$ G(x) = \cases{\dfrac{20}{\sqrt{21}} \arctan\left( \dfrac{\tan(x/2)}{\sqrt{21}}\right) & for $x < \pi$\cr
                 \dfrac{10 \pi}{\sqrt{21}}& for $x = \pi$\cr
\dfrac{20}{\sqrt{21}} \arctan\left( \dfrac{\tan(x/2)}{\sqrt{21}}\right) + \dfrac{20 \pi}{\sqrt{21}} & for $x > \pi$\cr}$$
Here is its graph:

And the correct integral is $G(2\pi) - G(0) = 20 \pi/\sqrt{21}$.

EDIT: Why do we get a formula for the  antiderivative that has a discontinuity where the integrand doesn't?  On a deeper level, what's going on here is that
the integrand has simple poles in the complex plane (at the zeros of $\cos(z) + 1.1$), and so any antiderivative will have branch cuts.  When a branch cut intersect the real line we get a discontinuity.

EDIT: following answer is incorrect. Disregard, trying to fix.

I don't know where you are getting that crazy antiderivative, but here is how I would do it.

$$\int \frac{dx}{1.1 + cos x}$$
$$x = \cos^{-1}u, \space \space dx = \frac{-1}{1-u^2}$$

$$-\int \frac{du}{(1.1 + u)(1+u)(1-u)}$$

and now we can use partial fracitonal decomposition.

$$\frac{A}{1.1+u}+\frac{B}{1+u}+\frac{C}{1-u} = \frac{1}{(1.1 + u)(1+u)(1-u)}$$

$$ A(1+u)(1-u) + B(1.1+u)(1-u) + C(1.1+u)(1+u) = 1$$

Skipping a few steps for the interest of time, $A = 4.7619, B=-5, C=0.238095$

Our integral is equal to $-4.7619 \ln |1.1 + u| +5 \ln |1+u| -0.238095 \ln|1-u|$

Plug in u = cos x.


Is $\iint f(x,y)\; dx\;dy$ called an iterated integral or iterated integrals In my Calculus 4 class we are learning about finding double and triple integrals. For example, one finds a double integral
$$
\iint_E f(x,y)\;dA
$$
by writing it as
$$
\int_a^b\int_c^d f(x,y)\; dx\;dy
$$
Is this call an iterated integral or is it iterated integrals? (So plural or singular.)

An excerpt from the text

  Stewart -- Calculus - Early Transcendentals, 6th Ed (2008)

  " . . . in this section we see how to express a double integral as an iterated integral, which can then be evaluated by calculating two single integrals."

On my course that I'm creating at the minute, I usually refer to them as double or triple integrals, i.e

Double Integral $$\iint_R dA$$
Triple (Volume) Integral $$\iiint_V dV$$

That way, it's easier to understand, it won't scare the crap out of your students. Otherwise I'd call it an iterated integral, since it's one big integral with several parts that is.
Hope this helps!


Let $a$, $b$, and $c$ be integers satisfying $a^2 + b^2 = c^2$. Prove: $abc$ must be even. I'm pretty sure that this can be proved by reductio ad absurdum, and have a proof for that. However, I'm not sure how to prove this using any other method of proof. It's my first time taking a course in logic and I'm not sure where to go with this.

Hint: Think about even and odd numbers. Can all 3 variables be odd?

Suppose $a$, $b$, and $c$ are odd. Then $a=2p-1$, $b=2q-1$, and $c=2r-1$. Inserting these into our formula, $a^2+b^2=c^2$, we get $(2p-1)^2+(2q-1)^2=(2r-1)^2$. Expanding, $4p^2-2p+1+4q^2-2q+1=4r^2-2r+1$, or $4p^2-2p+4q^2-2q+1=4r^2-2r$.

When written as $2(2p^2-p+2q^2-q+1)-1$, we see that the LHS is odd. Likewise, we see that the RHS is even. Because we can never have a value that is both even and odd, we get a contradiction. We can then say one of $a$, $b$, or $c$ must be even.

Let's say that $a$ is even; then $a=2n$, and $abc=2nbc$. We see that $abc$ is even. The other two cases are similar.

Therefore, we can say that $abc$ is even, in general.


Combinations Question regarding placement in multiple teams with restrictions? I've been having trouble with the following question:

  Nine players are to be divided into two teams of four, and one umpire. If two particular people cannot be on the same team, how many different combinations are possible?

The answer is 210, but I keep getting 70. The way I've been doing it is:

Let the three groups: the two teams and the umpire be represented as:

A _ _ _

B _ _ _

_

I have placed the three people in A's team: $7C3$ and B's team: $4C3$, then dividing by $2!$ due to the identical group sizes. What am I doing wrong?

Thanks.

Let's call the particular ones $A$ and $B$.

If none of the particular ones is chosen as umpire then there are
$7\times\binom{6}{3}=140$ possibilities ($1$ out of $7$ is picked
to be the umpire and then $3$ out of $6$ are chosen to join $A$)

If $A$ is chosen as umpire then there are $\binom{7}{3}=35$ possibilities
($3$ out of $8$ are picked to join $B$)

If $B$ is chosen as umpire then there are $\binom{7}{3}=35$ possibilities
($3$ out of $8$ are picked to join $A$)

This exact problem is explained in 2 parts.
Part 1: https://www.youtube.com/watch?v=LNX4Ya6Bg94 and
Part 2: https://www.youtube.com/watch?v=1qoX7RrMSNw&t=459s
The two players can be separated by making 1 an umpire.

So you have 140 (2 players on separate teams) +70 (player as umpire)


Prove that $\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$ by Taylor series. I want to prove that $\lim_{x\to 0}\frac{\ln (1+x)}{x}=1$ by Taylor series. I have get to the step: $$\lim_{x\to 0}\frac{\ln (1+x)}{x}=\lim_{x\to 0}\frac{x-\dfrac{x^2}{2}+\dfrac{x^3}{3}-\dfrac{x^4}{4}+\cdots}{x}$$

Next, when $x\neq0$, $\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots}{x}=\frac{1}{x}\left(x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots\right)=1-\frac{x}{2}+\frac{x^2}{3}-\cdots$. This is valid because $\sum_{n=1}^\infty(-1)^{n+1}\frac{x^n}{n}$ converges for every $x\neq 0$ near $0$, and then we can multiply a constant $\frac{1}{x}$ inside it. However, how can I rigorously and easily say that $\lim_{x\to 0}(1-\frac{x}{2}+\frac{x^2}{3}-\cdots)=1$?

You can be pretty brutal about this and just expand with the triangle inequality:

\begin{align*}
\left\lvert x - \frac{x^2}2 + \frac{x^3}3 - \cdots\right\rvert \le x + x^2 + x^3 + \cdots = \frac{x}{1 - x} \to 0
\end{align*}

as $x \to 0$.

That being said, it sorta misses the point to use Taylor series like this*. The limit you're looking at is the definition of
$$\left.\frac{d}{dx}\bigl(\ln(1 + x)\bigr)\right\vert_{x=0}$$
which you would need to compute in order to even know the Taylor series.

* If you've only defined log via Taylor series, or every used the derivative of log to compute the Taylor series.

$\dfrac{\log(1+x) -\log(1)}{x} =$

$ \log'(t) =1/t,$  $1< t<1+x.$

$\lim_{x \rightarrow 0} (1/t) =1.$

$\lim_{x \rightarrow 0} \dfrac{\log(1+x)-\log(x)}{x} = 1.$

Used: Mean Value Theorem.


Do we have $R\simeq S$ for two submodules $R,S$ of $A^n$? Let $A$ be a commutative ring with identity. Given two submodules $R,S$ of $A^n$ (where $n\in\Bbb N$), if there exists an isomorphism of $A$-modules $A^n/R\simeq A^n/S$, then do we have $R\simeq S$?

Note that this is definitely false for quotients of non-free modules: see, e.g., Quotient modules isomorphic $ \Rightarrow$ submodules isomorphic or Isomorphy of quotient modules implies isomorphy of submodules .

This is false even for direct summands. For example, take $R=\mathbb{R}[x,y,z]/(x^2+y^2+z^2=1)$, the co-ordinate ring of the real sphere. Let $P$ be defined as the kernel of the surjective map $R^3\to R$ given by $(x,y,z)$. ($P$ is the tangent bundle of the sphere). Then $P\oplus R\cong R^3$, but $P$ is not free.

No.

Let $A=\Bbb Z^\Bbb N=\{\,f\colon \Bbb N\to\Bbb Z\,\}$, $n=1$, $R=\Bbb Z=\{\,f\in A\mid \forall n>0\colon f(n)=0\,\}$, and $S=0$. Then $A^1/R\cong A^1/S\cong A$, but of coure $R\not\cong S$.


is strong topology on a metric space the topology that is induced by metric? is strong topology on a metric space the topology that is induced by metric? Is a open set in weak topology also open in strong topology?

Yes, in a linear space $X$ that is normed, e.g. we have a metric $d$ from the norm, and the strong topology is the one induced by $d$ (so generated by the $d$-balls). The weak topology looks at all continuous linear functions from $(X,d)$ to the underlying field (often the reals). This set is called $X^\ast$, the dual of $X$.

The weak topology $\mathcal{T}_w$ is the smallest topology that makes all functions in $X^\ast$ continuous. By definition the $d$-topology is one of those topologies, so by minimality, $\mathcal{T}_w \subseteq \mathcal{T}_d$. This is only an equality of topologies in the finite dimensional case. Otherwise the weak topology is strictly smaller (i.e. weaker), hence the name.

If we consider two topologies;The 1st topology is greater or bigger or stronger or finner whatever you says if it contains all the open sets of 2nd topology. So you are correct.


orthonormal basis of infinite dimensional Hilbert space H is not a basis of H as vector space? Apparently the orthonormal basis $(e_n)_{n\in \mathbb{N}}$ of the Hilbert space $H$ (in special case, infinitly dimensional) is not a basis of $H$ as a vectorspace. Is there a way to prove this?

Every vector space basis of an infinite dimensional Banach or Hilbert space is necessarily uncountable. This is actually a non-trivial result and follows from the Baire category theorem, see e.g. Let $X$ be an infinite dimensional Banach space. Prove that every Hamel basis of X is uncountable.

A basis means that for everu $u\in H$ you have $u=a_{i_1}e_{i_1}+...+a_{i_n}e_{i_n}$ for $n$ finite, and orthogonal basis means that $u=\sum_{n\geq 0}a_ne_n$ where the sequence $\sum_{i=0}^{i=n}a_ie_i$ converges, so for orthogonal basis, all the coefficients may not be zero.


Problems With Recursively Defined Sets I am having an issue with this problem.

  A  set, S,  of  positive  integers  is  defined  recursively  by  the  rule:

  $1 \in S$,
  If $n\in S$, then $2n-1 \in S$

  List  all  the  elements  in  the  set $S$.

The reason I am having difficulty is because plugging in $1$ into $2n-1$ provides me with the original statement, namely, that $1\in S$

This is far as I have been able to get with the problem and I would appreciate some helpful hints.

Ok, so all that that means is that $1$ is the only element in $S$; it's the only one we put in there as a 'base', and using the 'step', it turns out we can't get any others in there.

I think that your set  $S$ is simply not well-defined...

Taking $S_1 = \mathbb N - \{0\}$, this set satisfy the two conditions stated above.
But $S_2 = \{1\}$ too ! If such a set $S$ is well-defined via 1. & 2., it must be the only set satisfying those conditions, so $S_1 = S_2 = S$... contradiction...


Doubling the Sides of the Cube For the following question

    A cubical block of metal weighs 6 pounds.How much will another cube of the same metal weigh if its sides are twice as long ? Ans=48

Here is how I am solving it but I am getting 12 as the answer

For the current cubical block $12$ edges of cube weigh = 6 pounds

$1$ edge weighs = $\frac{6}{12} =  \frac{1}{2} pound$

Doubled edge would weigh $2 \times \frac {1}{2} = 1$ so 12 edges would weigh $12 \times 1 = 12$

HINT:

Write down the volume of a cube with edge length $x$. Think of an aquarium with all sides equal.
As 1. with length $2x$.
What's the ratio of the volumes?
How heavy is the larger cube?

8 edges of a cube. You have to do 6x8 in order to find your answer because you're doubling the lengths of each side. The logical answer may seem like 12, but the answer is actually 48


Finding solutions to $x^2+y^2+z^2=w^3$ Suppose $w$, $x$, $y$, and $z$ are all positive integers less than $100$. Find all such solutions to the equation $x^2+y^2+z^2=w^3$.

This problem was in a competition I participated in this past week, and I was unable to determine a clean way to solve it. Is there a more efficient way to approach it (perhaps using a variant of Legendre's three-square theorem?) without checking all values of $w$ less than $100$?

For the equation:

$$x^2+y^2+z^2=r^3$$

Will make a replacement that formula was compact.

$$c=2(q-p-s)t$$

$$d=s^2+t^2-q^2-p^2+2p(q-s)$$

$$k=p^2+t^2-q^2-s^2+2s(q-p)$$

$$n=p^2+t^2+s^2-q^2$$

$$j=p^2+s^2+t^2+q^2-2q(p+s)$$

$p,s,t,q$ - integers asked us. Then decisions can be recorded.

$$x=dn^2+2cnj-dj^2$$

$$y=cj^2+2dnj-cn^2$$

$$z=k(n^2+j^2)$$

$$r=n^2+j^2$$

You can write more compact.  You can write this:

$$x=3(p-k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$y=3(p-k+2t)(p^2+2k^2-2kt+2t^2)s^3$$

$$z=3(p+2k-t)(p^2+2k^2-2kt+2t^2)s^3$$

$$r=3(p^2+2k^2-2kt+2t^2)s^2$$

$p,k,t$ - integers asked us.

If there is a one simple solution, it will be necessary to reduce to the corresponding $s$.

There are other formulas. But they are bulky. Don't know whether it makes sense to write them.


Does fractional part converge in distribution to a uniform random variable? Let $X$ be a continuous random variable with a density function $f(x).$ Let $\{x\}$ denote the fractional part of a real number. I am tryng to prove that
$$
\mathbb{P}[\{nX\}\leq z] \rightarrow z, \ \ \forall z \in [0,1]
$$

Can anyone help me in this ?

My attempt:

If $g_n(z)$ is the density of $\{nX\}$, I got $g_n(z)=\sum_{m \in \mathbb{Z}}\frac{1}{n}f(\frac{m}{n}+\frac{z}{n})$. Can I justify from here ?

One can assume nice properties on $f$ if needed but I want to keep those assumptions as minimal as possible.

The obvious counterexample is the constant case $f(x) = \delta_{1/2}$ (for the Dirac $\delta$).  More generally, I doubt this is true of pathological continuous densities constructed, say, with peaks at all dyadic rational numbers.

If you specify extra smoothness then it may be true.


Calculating homology groups of a generic knot I am trying to calculate the homology groups of a knot embedded in $S^3$. This is what I have so far:
$$H_0(S^3-K)=\mathbb{Z}$$
since $S^3-K$ is path connected; from Alexander duality I have that:
$$H_n(S^3-K)=0 \qquad n\geq3$$
Again from Alexander duality I think it should be
$$H_2(S^3-K)\simeq H^0(K)=\mathbb{Z}$$
I need to find $H_1(S^3-K)$, but I do not know how. I was thinking to use Mayer-Vetoris method as follows:
$$0=H_1(S^3)\rightarrow H_1(S^3,S^3-K)\rightarrow H_1(S^3-K)\rightarrow H_1(S^3)=0$$
but I do not know how to find $H_1(S^3,S^3-K)$
EDIT I just thought I have to use Hurevicz theorem, so that $H_1(S^3-K)\simeq \pi_1(S^3-K)/[\pi_1,\pi_1]$

I am surprised that you know Alexander duality but not $H^1(S^1)$. However, you are right that Alexander duality shows
$$\tilde H_i(S^3 \setminus K) \approx \tilde H^{2-i}(K) \approx \tilde H^{2-i}(S^1). $$
Here $\tilde H_j$ and $\tilde H^j$ denote the reduced homology and cohomology groups. Except for $j = 0$ they agree with the unreduced groups. Moreover, if the reduced group in dimension $0$ is $0$, the unreduced group in dimension $0$ is $\mathbb Z$.
You correctly conclude that $H_i(S^3 \setminus K) = 0$ for $i \ge 3$.
The (reduced) cohomology groups of $S^1$ can be computed by exactly the same methods as the homology groups of $S^1$ which gives $\tilde H^1(S^1) = \mathbb Z$ and $\tilde H^i(S^1) = 0$ for $i \ne 1$. Let us come back to this point later. Anyway, we get
$$\tilde H_0(S^3 \setminus K) \approx  H^2(S^1) = 0, $$
$$H_1(S^3 \setminus K) \approx H^1(S^1) = \mathbb Z, $$
$$H_2(S^3 \setminus K) \approx \tilde H^0(S^1) = 0 . $$
Concerning the cohomology of $S^1$: We can either compute it as indicated above or use Alexander duality. In fact, consider $S^1$ embedded as the set $S = \{(x_1,x_2,0,0) \mid (x_1,x_2) \in S^1\}$. It is well-known that in this special case $S^3 \setminus S \simeq S^1$. Thus you know the homology of $S^3 \setminus S$ and via Alexander duality the cohomology of $S = S^1$.

Classical knots in 3-space have no interesting homology. Instead, Alexander looked at the homology of their 2-fold covering spaces (an easy to see invariant that distinguishes a lot of knots). Reidemeister showed, shortly thereafter, that linking numbers in non-cyclic coverings (just a bit harder to see) fill in all the known gaps (as revealed now through 19 crossings). -- Ken Perko, lbrtpl@gmail.com


Finding X from Exponential Equations $$2^x \cdot 4^{1-x}= 8^{-x}$$

I wrote all the base numbers as a power of 2 but I'm not sure what to do after.


what is the difference between exhaustive proof and proof by cases? I am currently reading Kenneth Rosen's book on Discrete Mathematics.
There I came across these two kind of proof strategy.

Proof By Cases
Exhaustive Proof

But I was not able to figure out the differences among them.
Here are the definitions mentioned in the book.

To prove a conditional statement of the form (p1 ∨ p2 ∨· · · ∨ pn) → q the tautology [(p1 ∨ p2 ∨ · · · ∨ pn) → q] ↔ [(p1 → q) ∧ (p2 → q) ∧ · · · ∧ (pn → q)] can be used as a rule of inference. This shows that the original conditional statement with a hypothesis made up of a disjunction of the propositions p1, p2, ... , pn can be proved by proving each of the n conditional statements pi → q, i = 1, 2, . . . , n, individually. Such an argument is called a proof by cases. Sometimes to prove that a conditional statement p → q is true, it is convenient to use a disjunction p1 ∨ p2 ∨ · · · ∨ pn instead of p as the hypothesis of the conditional statement, where p and p1 ∨ p2 ∨ · · · ∨ pn are equivalent.
EXHAUSTIVE PROOF : Some theorems can be proved by examining a relatively small number of examples. Such proofs are called exhaustive proofs, or proofs by exhaustion because these proofs proceed by exhausting all possibilities. An exhaustive proof is a special type of proof by cases where each case involves checking a single example.

Please provide an example along with the explanations to illustrate the differences.
Thank You.

Proof by cases tends to be about splitting your proposition into cases, at least some of which are general while an exhaustive proof looks at every case in a way that is not general. For example if one were to prove a property about natural numbers, it might be easier to say prove it for even numbers and then prove it for odd numbers separately. This would be proof by cases. On the other hand, if one were to prove a property about, say, all natural numbers less than 10 then one could just check the proposition for each of the natural numbers. There would be no generality and this would be an exhaustive proof.

Here are two examples, see if you can decide which method might apply to which.

The maximum diameter for a tree on 10 vertices where no vertex has degree 2 is 5
Any isometry of $\mathbb R^3$ can be represented as the product of up to three reflections.

In simple language what I understood is, Proof By Exhaustion is proving something for every case that could possibly occur whereas in Proof By Case numbers are split into categories and the statement is shown to be true for each category


How to know if N is normal to the plane or not? I have N = <1 , -2 , 3 > and the plane 2x - 4y + 6z = 5
How do i know if N is normal to the plane or not? please be detailed with the answer as i'm very lost.

Hint The normal of any plane $ax + by + cz = d$ is $(a,b,c)$, hence in this case $(2,-4,6)$.

Now, is your $N = (1, -2, 3)$ parallel to $(2, -4, 6)$?

In the equation for a plane the  coefficients of each of the coordinates tells you the direction that is perpendicular to the plane. You can tell if the vector given is perpendicular by measuring the angle between it and the perpendicular vector. How to do that is a matter of choice. I think easiest in this case is to note that 2*N is gives the coefficients so obviously N is parallel and thus perpendicular to the plane.


Need help solving a bi-quadratic polynomial.... The polynomial to be factorised as a product of two factors is- $$x^4+3x^2+6x+10$$. I checked the solution in wolfram alpha to be- $$(x^2-2x+5)(x^2+2x+2)$$. I tried to factorise it by expressing it as a sum of two squares $$(x^2+1)^2+(x+3)^2$$. But I cannot solve it. Please help. Thanks a lot in advance.

This is going beyond my knowledge in irreducible/reducible forms.. but it seems clear that if we assume at first glance that we have (which is not the way I would go about factoring forms like this)
$$
\left(x^2+ax\cdots\right)\left(x^2+bx\cdots\right)
$$
then we know that
$$
\left(x^2+ax\cdots\right)\left(x^2+bx\cdots\right) = x^4 +(a+b)x^3 + \cdots
$$
thus we must have $a+b = 0\implies a = -b$

but like I said I know there must be some group theory approach thanks to Galois et al.


Is it true that limit points of set of limit points don't exist? Is it true that limit points of set of limit points don't exist ?

I am not getting any example in which set of limit points has any limit point.

Perhaps the simplest example is the discrete topology, Let $X$={1,2,3,...,n}.We shall define discrete topology on it.Clearly each of $1,2,3,...,n$ are limits of {1},{2}, {3},..., {n} respectively.There collection  {1,2,...,n} has limit points $1,2,...,n$.


If $\int_{0}^{1}\dfrac{\sin{x}}{1+x}dx=b$, find $\int_{0}^{1}\dfrac{\cos{x}}{1+x}dx$ if we Assmue that $\int_{0}^{1}\dfrac{\sin{x}}{1+x}dx=b$,Find the value
$$\int_{0}^{1}\dfrac{\cos{x}}{1+x}dx$$  use the $b$ function expression

I try: let $$b=\int_{0}^{1}\dfrac{\sin{x}}{x+1}dx=\cos{1}\int_{1}^{2}\dfrac{\sin{x}}{x}dx-\sin{1}\int_{1}^{2}\dfrac{\cos{t}}{t}dt$$
and simaler we have
$$a=\int_{0}^{1}\dfrac{\cos{x}}{x+1}dx=\sin{1}\int_{1}^{2}\dfrac{\sin{t}}{t}dt+\cos{1}\int_{1}^{2}\dfrac{\cos{t}}{t}dt$$

$$\int_0^1\frac{\sin x}{1+x}dx=b$$
$$I=\int_0^1\frac{\cos x}{1+x}dx$$

$$I(t)=\int_0^1\frac{\cos(tx)}{1+x}dx$$
$$I'(t)=\int_0^1\frac{-\sin(tx)x}{1+x}dx=-\int_0^1\sin(tx)dx+\int_0^1\frac{\sin(tx)}{1+x}dx$$
$$J(t)=\int_0^1\frac{\sin(tx)}{1+x}dx$$
$$J'(t)=\int_0^1\frac{\cos(tx)x}{1+x}dx=\int_0^1\cos(tx)dx-\int_0^1\frac{\cos(tx)}{1+x}dx$$

$$I'(t)=-\left[\frac{\cos(tx)}{t}\right]_0^1+J(t)\Rightarrow I'(t)=\frac{1-\cos(t)}{t}+J(t)$$
$$J'(t)=\left[\frac{\sin(tx)}{t}\right]_0^1-I(t)\Rightarrow J'(t)=\frac{\sin(t)}{t}-I(t)$$

and we know that:
$$I(0)=\int_0^1\frac{1}{1+x}dx,J(0)=0,J(1)=b$$

If we differentiate we get:
$$I''(t)=\frac{d}{dt}\left[\frac{1-\cos(t)}{t}\right]+\frac{\sin(t)}{t}-I(t)$$
now rearrange to get:
$$I''+I=\frac{d}{dt}\left[\frac{1-\cos(t)}{t}\right]+\frac{\sin(t)}{t}$$

Consider $f(x)=\frac{e^{ix}}{x+1}$.  Using this, we can integrate $f(x)$ and take the real part to equal $\frac{\cos{x}}{x+1}$ and the imaginary
part to equal $\frac{\sin{x}}{x+1}$.

Now, integrate $f(x)$:
$$I=\int_0^1 \frac{e^{ix}}{x+1} \; dx$$ $$I=\int_1^2 \frac{e^{iu-i}}{u} \; dx$$ $$I=e^{-i} \int_i^{2i} \frac{e^t}{t} \; dt$$  $$I=e^{-i}\left(\text{Ei}\left(2i\right)-\text{Ei}\left(i\right)\right)$$

$$I\approx0.601044+0.284227i$$

Now equating $I$ to the original integral:
$$b= \int_0^1 \frac{\sin{x}}{x+1} \; dx =\operatorname{Im}\left(I\right)
 \approx 0.284227$$
Similarly:
$$\int_0^1 \frac{\cos{x}}{x+1} \; dx = \operatorname{Re}\left(I\right) \approx 0.601044\approx \boxed{2.11466\cdot b}$$


Convergence of $\sum\limits_{n=1}^{\infty} {\dfrac{(-1)^{n-1}} {(\sqrt{n}+(-1)^{n-1})^p}}$ Find $p$ that makes $\sum\limits_{n=1}^{\infty} {\dfrac{(-1)^{n-1}}{(\sqrt{n}+(-1)^{n-1})^p}}$ converge. Which $p$ makes the series converge absolutely?

I think that it converges for $p>0$, can I use: ${\dfrac{1}{{{{\left( {\sqrt{n} + {{\left( { - 1} \right)}^{n - 1}}} \right)}^p}}}} \sim \dfrac{1}{{{n^{\frac{p}{2}}}}}$ to conclude the series converges absolutely for $p>2$?

Thanks in advance!

Answered in the comments, but let's do it explicitly.

$\sum\limits_{n=1}^\infty a_n$ converges if and only if $\lim\limits_{n\to\infty}a_n=0$ and $\sum\limits_{n=1}^\infty(a_{2n-1}+a_{2n})$ converges.

For $a_n=(-1)^{n-1}\big(\sqrt{n}+(-1)^{n-1}\big)^{-p}$, this translates into [$p>0$ and] $\color{blue}{p>1}$, since $$a_{2n-1}+a_{2n}=f(1/\sqrt{2n}),\quad f(x):=x^p\big((\sqrt{1-x^2}+x)^{-p}-(1-x)^{-p}\big),$$ so that $\lim\limits_{x\to0^+}x^{-p-1}f(x)=-2p$, and we're done by the comparison test, with $\sum\limits_{n=1}^\infty n^{-(p+1)/2}$.

For $n\geq 1$, let $u_n=\frac{(-1)^{n-1}}{(\sqrt{n}+(-1)^{n-1})^p}$

1) For $p<0$, we have

$u_n=(\sqrt{n})^{-p}(-1)^{n-1}\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^{-p}$

So $u_n \sim (-1)^{n-1}(\sqrt{n})^{-p}$ ($n\rightarrow +\infty$)

so $\lim_{n\rightarrow +\infty} |u_n|=+\infty$

Thus $(u_n)$ doesn't tend to 0.

The series is divergent.

2) For $p=0$, $u_n=(-1)^{n-1}$

We know that the series is divegent.

3) Suppose $p>0$

We have $u_n=\frac{(-1)^{n-1}}{(\sqrt{n})^p}.\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}$

We deduce that

$|u_n| \sim \frac{1}{n^{p/2}}$ ($n\rightarrow +\infty$)

So the series $\sum u_n$ is absolutely convergent if and only if $p/2>1$, if and only if $p>2$.

But absolutely convergent implies convergent

4) Suppose $0<p\leq 2$

We make a series expansion for $n\geq 2$

$\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}=\sum_{k=0}^{+\infty} {-p \choose k}\left(\frac{(-1)^{n-1}}{\sqrt{n}}\right)^k$

Multiplying by $\frac{(-1)^{n-1}}{n^{p/2}}$, the power over $n$ is $\frac{p}{2}
+\frac{k}{2}=\frac{k+p}{2}$

and $\frac{k+p}{2}>1\Leftrightarrow k>2-p$

So for $k\geq 2$ there will be absolute convergence

$\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}=1-p\frac{(-1)^{n-1}}{\sqrt{n}}+O\left(\frac{1}{n}\right)$

$\frac{(-1)^{n-1}}{(\sqrt{n})^p}.\frac{1}{\left(1+\frac{(-1)^{n-1}}{\sqrt{n}}\right)^p}=\frac{(-1)^{n-1}}{n^{p/2}}-p\frac{1}{n^{(p+1)/2}}+O\left(\frac{1}{n^{p/2+1}}\right)$

The last term is of the form $\frac{b_n}{n^{p/2+1}}$ with $(b_n)$ a sequence bounded by a real number $M$. However, the series $\sum \frac{M}{n^{p/2+1}}$ is absolutely convergent because $p/2+1>1$, so $\sum \frac{b_n}{n^{p/2+1}}$ is absolutely convergent.

The series $\frac{(-1)^{n-1}}{n^{p/2}}$ is an alternate Riemman series with $p/2>0$ so is convergent.

Finally the Riemman series $\sum \frac{1}{n^{(p+1)/2}}$ with positive terms is convergent if and only if $(p+1)/2>1$ i.e. $p>1$.

We can conclude that the series $\sum u_n$ is convergent if $1<p\leq 2$ as sum of convergent series.

and divergent if $0<p\leq 1$ as the sum of a divergent series and others convergent.


Exact value of trig functions Can we find the exact value(not numerical/approximation) of sin 1?
I tried to do so by solving a cubic equation using Cardano formula but I ended up with complex nested radicals): I was told to use other methods but that won't yield exact values

sin(1) is a Transcendental number.
http://en.wikipedia.org/wiki/Transcendental_number
http://en.wikipedia.org/wiki/Lindemann%E2%80%93Weierstrass_theorem

ie ,sin(1) is not a solution  a non-zero polynomial equation with rational coefficients.In other words sin(1) can't be expressed in closed form only using fractions and radicals.
Note: sin(a), where a is algebraic(not transcendental) ,is transcendental.

Does that help you?

Yes, we can. However the radicals will be nested as well as repeating. Using Precise-Rewritten method, exact value of Sin 1 degree is:
 $$√(2 - √(2  + √(2  \overline {+ √(2 + √(2  + √(2 + √(2 - √(2 - √(2 - √(2 + √(2 - √(2  - √(2  + √(2  - √(2}]/2$$

Please care the overline (not fraction) and closing brackets have collapsed by']'.

This method gives exact values of Sine of all integer angles. Please visit https://archive.org/details/ExactTrigonometryTableForAllAnglesFinal for the list.


Show $\sum_{n=1}^{k} 2^{a_n} = 2^{a_1+1}$ holds for all non negative, monotonic decreasing sequences. Let $(a_n)_{n\in\mathbb{N}}$ be a non negative, monotonic decreasing sequence with each $a$ being a whole number. Prove that there exists a $k$ such that
$$
\sum\limits_{n=1}^{k} 2^{a_n} = 2^{a_1+1}
$$
holds.

This result seems rather obvious and intuitive but I want a rather simple and concise proof, which I was so far unable to find.

Suppose otherwise. Since $2^{a_n}\geq1$, we could take $k$ such that $$\sum_{n=1}^k2^{a_n}<2^{a_1+1}<\sum_{n=1}^{k+1}2^{a_n}\Leftrightarrow$$ $$\sum_{n=1}^k2^{a_n-a_k}<2^{a_1-a_k+1}<\sum_{n=1}^{k+1}2^{a_n-a_k}.$$
Since the numbers in the last inequality are all integers, this must imply that $$\sum_{n=1}^{k+1}2^{a_n-a_k}-\sum_{n=1}^k2^{a_n-a_k}\geq2\Leftrightarrow$$ $$2^{a_{k+1}-a_k}\geq2.$$ However, since $a_{k+1}\leq a_k$, this is absurd. We've reached our contradiction.

If $a_n$ is allowed to be negative this isn't so. Assuming "whole number" means $a_n\ge0$:

Since $2^{a_n}\ge1$ there exists $k$ with  $
\sum\limits_{n=1}^{k} 2^{a_n} \ge 2^{a_1+1}.
$ Choose $k$ as small as possible satisfying this inequality, so $$
\sum\limits_{n=1}^{k} 2^{a_n} \ge 2^{a_1+1}
$$while $$
\sum\limits_{n=1}^{k-1} 2^{a_n} < 2^{a_1+1}.
$$

Now say $$2^{a_1+1}=Ka_{k}.$$Since $
\sum\limits_{n=1}^{k-1} 2^{a_n}
$is a multiple of $a_k$ we have $$
\sum\limits_{n=1}^{k-1} 2^{a_n} \le (K-1)a_k.
$$It follows that in fact $\sum\limits_{n=1}^{k-1} 2^{a_n} = (K-1)a_k$, since otherwise $\sum\limits_{n=1}^{k-1} 2^{a_n} \le (K-2)a_k$, which would imply $$\sum\limits_{n=1}^{k-1} 2^{a_n} \le (K-1)a_k<2^{a_1+1},$$contradiction. So $$\sum\limits_{n=1}^{k} 2^{a_n} = Ka_k=2^{a_1+1}.$$


A question related to Wave Equation Let $L>0$. Suppose $f, g$ are $C^2$ functions on $\mathbb{R}$ such that
$$f(t)+f(-t)+\int_{-t}^t g(s)\,ds=0$$ and
$$f(L+t)+f(L-t)+\int_{L-t}^{L+t} g(s)\,ds=0$$
for all $t\in \mathbb{R}.$
Does it follow that $f, g$ are odd periodic functions of period $2L$?

I think so.

What you've written is the D'Alembert formula for the wave equation in 1 dimension on the line, specifying that $u(0,t) = u_t(0,t) = u(L,t) = u_t(L,t) = 0$ for all times $t$.

The generic Fourier mode to this equation is $e^{i\xi (x -ct)}$, and the representation of the solution in terms of the Fourier inversion formula (in the sense of tempered distributions if needed) is basically that you are integrating over all frequencies $\xi$. But in this case, there can be no contribution from any mode except where $\xi$ is $\frac{n\pi}{L}$ because of your imposed conditions, and indeed only the sine portion of that term. This basically gives you the conclusion you sought for $f$ and $g$.

I gather from the negative vote that I was too vague. I will try again with a less hand-wavy method, without reference to the Fourier theory.

You can consider that in general, on the line, the solution to the wave equation is the superposition of a wave travelling to the right and a wave travelling to the left. So if we set $c=1$, we have
$$u(x,t) = F(x+t) + G(x-t) = \frac{1}{2} \left(f(x+t) + f(x-t) + \int_{-t}^t g(s) ds \right)$$
where $f(x) = F(x) + G(x)$ and $g(x) = F'(x) - G'(x)$ (for the details of this transformation, see any textbook on elementary PDE, or the Wikipedia article on d'Alembert's formula).

The conditions translate to a statement that $F(t) + G(-t)= 0$, and that $F(L+t) + G(L-t) = 0$. Plugging the first equality into the second, we get $F(L+t) = F(-L+t)$, which is precisely the statement that $F$ is periodic of period $2L$. If we perform the substitution with respect to $G$ instead, we get that $G(L-t) = G(-L-t)$ for all values $t$, so $G$ is periodic of period $2L$ as well. Since $F$ and $G$ are $2L$ periodic, so are $f$ and $g$.

To see that $f$ and $g$ are odd, consider what happens when take a look at
$$f(-x) = F(-x) + G(-x) = -G(x) + (-F(x)) = - f(x)$$
This uses only the first condition above. Similarly, differentiate the first condition to get that $F'(t) - G'(-t) = 0$, and then substitute to get
$$g(-x) = F'(-x) - G'(-x) = G'(x) - F'(x) = - g(x)$$

There is an obvious counterexample.

When $f(t)=g(t)=0$ , they obviously satisfly $f(t)+f(-t)+\int_{-t}^tg(s)~ds=0$ and $f(L+t)+f(L-t)+\int_{L-t}^{L+t}g(s)~ds=0$ , but $f(t)$ and $g(t)$ should not belongs to odd periodic functions of period $2L$ .


Is there a sign for 'not less than', 'not greater than', etc.? I was wondering about this, just now, because I was trying to write something like:
$880$ is not greater than $950$.
I am wondering this because there is a 'not equal to': $\not=$
Not equal to is an accepted mathematical symbol - so would this be acceptable: $\not>$?
I was searching around but I couldn't find any qualified sites that would point me in that direction.

So, I would like to know if there are symbols for, not greater, less than, less than or equal to, greater than or equal to x.

Thanks for your help and time!

To answer the question, yes.
$$
a \nless b\\
a \ngtr b\\
a \nleq b\qquad a \nleqq b\qquad a \nleqslant b\\
a \ngeq b\qquad a \ngeqq b\qquad a \ngeqslant b
$$
and so on for many other mathematical relations
$$
a \nleftarrow b\\
a \nLeftarrow b\\
A \nsupseteqq B\\
A \nvdash \phi\qquad A \nVdash \phi\\
\nexists x
$$

I would just like to make it clear that ≮ is NOT the same as ≥

Here is an example:

1+²≮(1+)² is clearly not the same as 1+²≥(1+)²

Think of =-0.5 and =2 as examples to highlight this, because although when =-0.5, 1+²≥(1+)² but when =2, 1+²≱(1+)²

Therefore, think of ≮ as meaning "not greater than" and ≥ meaning "more than or equal to" but remember that they are not the same!


A plan to defeat a betting game where the odds of winning are 50/50. Help me understand why it's flawed. My friend has this plan where he implies that it's impossible to lose, as long as the odds of winning are 50/50 on each bet. His idea is that basically you keep doubling your bet until you win and then start over again.

So for example, you bet 1 dollar and you lose, your net profit is now -1 dollar. Now you double your bet to 2 dollars and you lose again so your net profit is -3 dollars. Now you double your bet to 4 dollars and you win. This means you gain 4 dollars and now your net profit is 1 dollar. So you've made a profit. Now you start again. The reasoning here being that it is highly unlikely for you to lose a 50/50 toss x number of times in a row.

My counter-argument here is that basically if you go in with 50 dollars with the aim of doubling up to 100 dollars, you have the same odds of winning if you do one bet of 50 dollars or the technique outlined above. I cannot wrap my head around explaining this issue in a clear manner though, so maybe you wonderful folk at Mathematics can help!

Oh and I've pointed out that he uses gamblers fallacy in very obscure way, as he insists you need to go back to betting 1 dollar once you've won. This appears to be an obscure case of gamblers fallacy to me as it implies there is some hidden force which are changing the odds on each individual coin toss.

Let's play out your specific scenario: you start with \$50 and want to stop at \$100.  That means you need to play your sequence (double until you win) 50 times without busting.

However, if your sequence begins with six losses, then you are busted, because you just lost $1+2+4+8+16+32=63$ dollars, and your system requires you to have \$64 left to bet.  Six losses in a row happens $1/2^6=\frac{1}{64}$ of the time.

To avoid 6 losses in a row, happens $\frac{63}{64}$ of the time, and you need to do this 50 times, so $(\frac{63}{64})^{50}\approx 0.455$.  Hence, if all you need to do is avoid six losses, there is about a 45% chance of "winning", i.e. getting to \$100.

However, there are other circumstances when you bust; at the beginning even five losses in a row will bust you, since $1+2+4+8+16=31$ dollars, and you need to bet \$32.  Hence until you get to \$63, you need to avoid five losses in a row.  Thus, the true answer is $$\left(\frac{31}{32}\right)^{13}\left(\frac{63}{64}\right)^{37}\approx 0.370$$
Thus your friend's system has a 37% chance of getting to \$100, and a 63% chance of getting to \$0.  You are much better off just betting the whole \$50 in a single bet.

The simple solution would be to sit and wait until 5 loses come. This will take statistically 32 turns. Then use this betting system (Assuming you lose) £5 £12 £30 £60 £70. At this point you have a 1/1024 chance of losing. £150 £200 £300 £500 £700. At this point you have a seriously low chance of this happening and when it does you'd be extremely likely to be rich already. You've risked £2027 however the chances that your previous winnings will cover this loss are 100.000s in your favour. Always stay with the same strategy of playing the game you're playing. For example: rollette. If you start on black stay on black.

With your friends strategy you will always be playing to gain £1 profit. By his tenth loss he has put the bet on of 512 losing £1023 for that £1 profit WHAT THE HELL IS WRONG WITH HIS LOGIC. It gets worse by his 15th hand he would be placing £16384 and has already lost 16383 so when that fails he has lost 32767 all for chasing the £1 profit. This is just plain stupid. Casinos have minimums and maximum betings for a reason and that is to put them in favour to win.

Now I'm not encouraging ANYONE to gamble but if you're going to do it regardless then please use mine over the original strategy. Casinos and so on are winners for a reason! Remember that. Bye for now and if you want to email me your thoughts please email andytomkins11@gmail.com p.s, my maths maybe slightly out as I done them in my head.


How was the quadratic formula found and proven? Possible Duplicate:
  Why can ALL quadratic equations be solved by the quadratic formula?
  History of Quadratic Formula

How was the quadratic formula $\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ found and proven?

I don't know for sure, but I imagine it was first found by thinking about rearranging squares and rectangles, as in the following image:

by completing the square. $a\neq 0$,  $ax^2+bx+c=a\left(x^2+\dfrac{b}{a}x+\dfrac{c}{a}\right)=\cdots$


Prove $\tan \theta = 3 \tan \alpha$ Prove $\tan \theta = 3 \tan \alpha$

The question is related to this diagram.

I tried approaching this problem using the cosine and sine rules for the triangles in this figure. I was hoping that if I could find $\sin \theta, \sin \alpha, \cos \theta$ and $\cos \alpha$, I could try to simplify $ \tan \theta$ to get $3\tan \alpha$. However, that didn't work. I ended up with

$\tan \theta = \frac{-BD^2}{BC^2}\cdot  \sin 2\theta $, which doesn't seem to be helpful. Any hints or answers would be appreciated.

It is obvious that $\bigtriangleup BDC$ is isoscales triangle. So, if you mark the midpoint of the side $\overline{BC}$ with $E$, you will have right triangle $\bigtriangleup BDE$, from which you can find: $$\tan \theta = \frac{|\overline{DE}|}{|\overline{BE}|}.$$

Next, your triangle $\bigtriangleup ADE$ is also right triangle, so it holds:
$$\tan \alpha = \frac{|\overline{DE}|}{|\overline{AE}|}.$$

Since $\quad \overline{AE} = 3\overline{BE} \quad$ we have:
$$\tan \alpha = \frac{|\overline{DE}|}{3|\overline{BE}|}.$$

Finally, multiplying with $3$ will result with:
$$3 \tan \alpha = \tan \theta.$$

By the law of sines for $\Delta ADC$ we obtain: $$\frac{AC}{\sin(\alpha+\theta)}=\frac{DC}{\sin\alpha}.$$
Also, $$AC=2BC=4DC\cos\theta.$$
Thus, $$4\sin\alpha\cos\theta=\sin(\alpha+\theta)$$ or
$$4\sin\alpha\cos\theta=\sin\alpha\cos\theta+\cos\alpha\sin\theta$$ or

$$3\sin\alpha\cos\theta=\cos\alpha\sin\theta$$ or
$$3\tan\alpha=\tan\theta.$$


The supremum value of $x^{2}y^{2}(x^{2}+y^{2})$ when $x+y=2n$ for some fixed $n\in \mathbb N $ Let  $S$  be  the  set  of  all  tuples $(x,y)$  such  that $x+y=2n$    for  a  fixed  $n\in \mathbb N$. Then what  is  the  supremum  value of   $x^{2}y^{2}(x^{2}+y^{2})$ $?$

I  substituted $y=2n-x$ $\ $ in $\ $ $x^{2}y^{2}(x^{2}+y^{2})$ and  naming  it  $f(x)$  tried  differentiating  so  at $f'(x)=0$  the  supremum  value . I  hope  that  method  was  correct. But  the  differentiation  was  too tedious and since  this  problem  was  set  in  an  exam  for  $2$  marks  only  I  have  doubt  that  there  is  some  shorter  trick to  get it . Is  there $?$ Please  let  me  know.

Thanks  for  any  help.

$$f(x,y)=x^2y^2\{(x+y)^2-2xy\}=4n^2t^2-2t^3=g(t)$$ where $t=xy$

Use Second derivative test

and honour $4xy\le(x+y)^2=(2n)^2$

$x_m = n + m, \quad y_m = n - m, \quad m \in \mathbb{N} $,  then $x_m + y_m = 2n$ and $x_m^{2}y_m^{2}(x_m^{2}+y_m^{2})$ goes to infinite when $m \rightarrow \infty $. Therefore there is no supremum value.


Check that two function $f(x,y)$ and $g(x,y)$ are identical Given that $f(x)$ and $g(x)$ are two polynomials of degree $n$, we know that if we can find $n+1$ distinct numbers $x_i$, $i=1,\cdots,n+1$ such that $f(x_i)=g(x_i)$ then $f(x)$ and $g(x)$ are identical.

However I don't know how to do this with two variable functions. How can I check if a function $f(x,y)$ has the form $g(x,y) = Ax^2+Bxy+Cy^2+Dx+Ey+F$ or not? How many pairs $(x_i,y_i)$ do I have to check to make sure that $f(x,y)$ is identical to $g(x,y)$?

Regards.

If $f$ and $g$ are two polynomials of degree $n$, and there are distinct points $x_1, \dots, x_{n+1}$ such that $f(x_i) = g(x_i)$ for $i = 1, \dots, n+1$, then $f = g$. One way to see this is that a generic degree $n$ polynomial has the form

$$a_nx^n + a_{n-1}x^{n-1} + \dots + a_1x + a_0$$

which contains $n+1$ coefficients. Given $n+1$ points $(x_i, y_i)$ which satisfy the polynomial equation, we obtain $n+1$ equations in $n+1$ unknowns which uniquely determine the coefficients, and hence the polynomial.

If now $f$ and $g$ are polynomials in two variables of degree $n$. A generic such polynomial has the form

$$\sum_{i=0}^n\sum_{j=0}^ia_{ij}x^jy^{i-j}$$

which contains $N:=\frac{(n+1)(n+2)}{2}$ coefficients. Given the situation for one variable polynomials, you may guess that if there are distinct points $(x_1, y_1), \dots, (x_N, y_N)$ with $f(x_i, y_i) = g(x_i, y_i)$ for $i = 1, \dots, N$, then $f = g$. Your guess would be wrong. While it is true that $N$ points of agreement is enough to show that $f = g$, it is not true that any $N$ points will do. Each point will still give an equation so that you obtain $N$ equations in $N$ unknowns, you might get some redundancy which means the system will not have a unique solution.

Here is the basic idea.

Same idea for polynomial in two variables. See how many coefficients they have. This is just a matter of counting. Let us say this number is $N$. Then if the polynomials match at $N$ distinct points which meet some requirements (to be explored below), then they have to be equal.

If the degree is 1, then N=3.
If degree is 2 then N=6
and so on.

Note, if you take the view of the coefficients, then the value of the polynomial at some point gives a linear equation in the coefficients. So if the $N$ points are chosen so that these are linearly independent equations, then if two polynomials match at these points, then they are equal.

I have found that choosing the grid points, so that no more than 2 are in a straight line works (I don't have a proof for this).

** Based on comment from  Michael Albanese**

I want to emphasize, that I do not have a proof and I hope someone would improve my answer. This is just the basic idea.


Why do we say "radius" of convergence? In an intuitive sense, I have never understood why a power series centered on $c$ cannot converge for some interval like $(c-3,c+2]$.

Also, I have had a few professors casually mention that a series converges for a disk in the complex plane, centered on $c$ and with the radius of convergence as its radius.

Is this just a deep result that I havent seen yet because I havent taken enough real/complex analysis? Or is there an obvious reason for this.

Edit: perhaps a better way to ask my question: why is it that, if the series converges when $x$ is between $c$ and $c+a$, then it also converges for $x$ between $c$ and $c-a$

While it is true that in complex analysis, power series converges on discs (hence the name 'radius of convergence'), this is not necessary to see why real power series converge on a symmetric interval about their centre.

A power series with real coefficients centred at the point $c$ can be written as
$$
\sum_{n=0}^{\infty} a_n (x-c)^n,
$$
and it will converge whenever we plug in a value of $x$ so that the resulting series of real numbers converges. If $b = |x-c|$, then we're really examining the convergence of the series of numbers
$$
\sum_{n=0}^{\infty} a_n b^n.
$$
(This statement is precisely the ratio test!)
This will converge (say) when $0 \leq b < R$, so we are allowed to substitute in values of $x$ satisfying $0 \leq |x-c| < R$. Therefore, the values of $x$ for which the series $\sum_{n=0}^{\infty} a_n (x-c)^n$ are those in the interval $(c-R, c+ R)$. (One must, of course, check what happens when $x = c- R$ or $x = c+R$, as we could have convergence at one or both of the endpoints, but in any case we still get a symmetric interval about $c$.)


Solve $B \ln y +A y \ln y + A y-A =0$ for $y$ I would like to know if there is a (preferably closed-form) solution for

$B \ln y +A y \ln y + A y- A =0$ for $y$

Where $A, B \in \mathbb{R}^{+}$. I have reasons to think there isn't a closed form solution or an approximation that is valid when $r$ is in the order of $B$ and $B \rightarrow 0$.

Is there any theorem saying there are no closed-form solutions to this equation? If there is, do you think an approximate solution might be possible? If there isn't, any hint how to solve it?

Note: My question is motivated by the need for a solution to this problem:

$1-\frac{A}{x} \gamma (2,\frac{x}{B}) = 0$ for $x$

The solution needn't be exact, but it'd be great if it was in closed form (I'm trying to avoid numerical approximations). Especially, I'm interested in the case $r$ around $B$, but also would like convergence for $B \rightarrow 0$.

My approach was to perform an asymptotic expansion on the incomplete lower gamma function following http://dlmf.nist.gov/8.11, truncating the series after 2 terms. After simplifying, Mathematica gives me:

$x + x \frac{A}{B} e^{-\frac{x}{B}}+A e^{-\frac{x}{B}} -A =0$

So, after the substitution $x=B \ln y$, I get:
$B \ln y +A y \ln y + A y-A =0$ for $y$

Maybe there is a better approach to this problem?

Thanks!

Since $A>0$ (the only solution for $A=0$ would be $B=0$ and $y$ arbitrary anyway) we can divide by $A$ to obtain
$$(y+\frac{B}{A})\ln y = 1-y$$
Note that $y=1$ is certainly a solution to your equation. Furthermore, note that for $y<1$ the left hand side is negative, whereas the right hand side is positive and conversely, for $y>1$ the left hand side is positive, whereas the right hand side is negative.
Hence $y=1$ is the only solution to your equation.

I doubt that the equation has a nontrivial solution. Implicitly differentiating once,

$$
B/y + 2A + A\ln y = 0.
$$

Rearranging,

$$
\mathrm{e}^2y = \mathrm{e}^{-B/Ay}.
$$

Implicitly differentiating once more,

$$
\mathrm{e}^2 = -\frac{B}{A}\mathrm{e}^{-B/Ay}
$$

which seems to suggest that $y$ is constant.


Subsets $B$ of bounded subinterval $I$ is lebesgue measurable iff $\lambda^*(I)=\lambda^*(B)+\lambda^*(I\cap B^c)$ Hi I was reading Cohn's book and I have problem with the following exercises (only the return of b is what I don't know), I'd appreciate any help and suggestion, if necessary, for a):

  a) Show that a subset $B$ of $\bf{R}$ is Lebesgue measurable iff $\lambda^*(I)=\lambda^*(I\cap B)+\lambda^*(I\cap B^c)$ for any open interval.

  b) Let $I$ a bounded subinterval of $\bf{R}$. Show that a subset $B$ of $I$ is Lebesgue measurable iff it satisfies $\lambda^*(I)=\lambda^*(B)+\lambda^*(I\cap B^c)$

a) We only show the sufficiently. Let $A\subset \bf{R}$ and we may assume that $\lambda^* (A)<+\infty$. Let $\{(a_n,b_n)\}$ a sequence of open intervals such that $A\subset \bigcup_n(a_n,b_n)$ and $\sum_nb_n-a_n<\lambda^* (A)+\varepsilon$, where $\varepsilon$ is an arbitrary positive number. Thus

\begin{align}\lambda^*(A\cap B)+\lambda^*(A\cap B^c)\le \sum _n\lambda^*((a_n,b_n)\cap B)+\sum_n\lambda^*((a_n,b_n)\cap B^c)\\
= \sum _n\lambda^*((a_n,b_n)\cap B)+\lambda^*((a_n,b_n)\cap B^c)\\=\sum _n\lambda^*((a_n,b_n))=\sum_nb_n-a_n\\<\lambda^* (A)+\varepsilon\end{align}

Letting $\varepsilon \downarrow0$, $\lambda^*(A\cap B)+\lambda^*(A\cap B^c)\le \lambda^* (A)$. Hence $B$ is Lebesgue measurable.

(b) One side is obvious, but the problem is with the return...

First, show that if $B$ is a subset of a bounded closed interval $I$, then $\lambda_*(B)+\lambda^*(I\backslash B)=\lambda(I)$. Then, by assumption, we have that
$\lambda_*(B)+\lambda^*(I \backslash B)=\lambda^*(B)+\lambda^*(I\backslash B)$, hence, $\lambda_*(B)=\lambda^*(B)<\infty$ and $B$ is measurable.

We prove that the equation holds for each open interval and then use the conclusion of a). Let $L$ be an open interval and $L=(L\cap I)\cup (L\cap I^{c})$. Since $I$ is measurable, we have $$\lambda ^{*}(L)=\lambda ^{*}(L\cap I)+\lambda ^{*}(L\cap I^{c})$$
Also, we have $$\lambda ^{*}(I\cap L)\leq \lambda ^{*}(B\cap I\cap L)+\lambda ^{*}(B^{c}\cap I\cap L)(1)$$ $$\lambda ^{*}(I\cap L^{c})\leq \lambda ^{*}(B\cap I\cap L^{c})+\lambda ^{*}(B^{c}\cap I\cap L^{c})(2)$$
Add these two inequalities and use the fact that $L$ is measurable, we have
$$\lambda ^{*}(I)=\lambda ^{*}(L^{c}\cap I)+\lambda ^{*}(L\cap I)\leq \lambda ^{*}(B)+\lambda ^{*}(B^{c}\cap I)=\lambda ^{*}(I)$$
This indicates that the inequalities in $(1)$ and $(2)$ should be strict equalities. Besides, it is easy to check, if an interval $E\cap I=\varnothing $, we have $$\lambda ^{*}(E)=\lambda ^{*}(E\cap B)+\lambda ^{*}(E\cap B^{c})$$Now, we have$$\lambda ^{*}(L)=\lambda ^{*}(L\cap I)+\lambda ^{*}(L\cap I^{c})=\lambda ^{*}(L\cap I\cap B)+\lambda ^{*}(L\cap I\cap B^{c})+\lambda ^{*}(L\cap I^{c}\cap B)+\lambda ^{*}(L\cap I^{c}\cap B^{c})$$ The first term is $\lambda ^{*}(L\cap B)$ and the third term is zero. The sum of second and fourth terms is $\lambda ^{*}(L\cap B^{c})$ since $I$ is measurable.


Possible Class equation for a group Determine the possible class equation for a group of order 21?

Until now I have found the following:

$1+3+3+7+7$

$1+1+1+3+3+3+9$

$1+1+1+1+1+1+1+7+7$

$1+1+1+1+1+1+1+1+1+3+3+3+3$

$1+1+1+\cdots +1 \ (21 \ \text{times})$

Is there any way to eliminate the choices from this equation? More importantly, how would we know that this is a complete list (Until now my attempt has just been guess and check after I found possible occurences of 1's) Is there any easier way to determine the class equation?

Hint: the number of 1's is the order of $Z(G)$, the center of $G$. Also, $|G/Z(G)|$ can not be a prime number. And of course $|Z(G)|$ divides the order of $G$. This leaves you with the first and last possibilities. In the last case the group is abelian.

I think $G$ is abelian. So $|Z(G)|=21$. Hence class equation is all of $1$'s.


How do I find the averge value of the function g(x) on the interval [1, + infinity]? I know how to find the average value of functions on an interval, but I am having trouble finding the average value of this function with the interval going to positive infinity. I tried doing what the hint says, but I'm not sure if I am on the right track. Can anyone suggest anything? Thank You!

Given the amount of your homework you have posted in the past four hours...

Hint: $u = \frac{\pi}{x}$.


Do we have to define natural numbers in geometry? I have been thinking about the axiomatization of geometry, and I don't know one thing.
Imagine you are defining triangles:

  Definition [Triangle]: A triangle is a polygon with 3 sides.

In this case, is it required to define $3$ (and, by extension, $\mathbb{N}$)?

Edit: Maybe I should explain myself more clearly. What I am doing is the following: I have the definition of triangle, now what do I have to define previously so that the definition can be completely understood? In this case, it is clear that I must define what a polygon is and what is a side/vertex of a polygon. I'm wondering whether I have to define "3", but I am not complaining specifically with number 3, it just happens that it is the number I needed to use in this case.

In order to write down definitions and axioms (without which those definitions don't really make sense), you need a logic - that is, a formal language in which you can write these statements.

The standard example is first-order logic, although there are others. And Euclidean geometry is easily formulated in first-order logic.

Now, the natural numbers are implicit in first-order logic! Specifically, defining what a sentence is, and what a proof is, requires us to already understand what a natural number is. And there's really no way around this, and nothing special about first-order logic in this regard (indeed, other logics tend to require more mathematical background - e.g. second-order logic basically requires all of set theory!).

So in order to have a context in which your definitions and axioms can be expressed, and theorems can be proved, you need to already have the natural numbers "in the background".

To clarify: it's not that we have to define "$3$" in order to express "A triangle is a polygon with $3$ sides," it's that we already need to understand $\mathbb{N}$ in order to be able to formulate the language within which this definition is being expressed.

This point of view can be seen in more detailed fashion at the answers to this question.

Without attempting to rule on the question I would like to express my opinion that there are two first order logics. There is the naive logic we define using the english language an all that it entails (though I think numbers dont play a big role). There is a second formalized FOL defined within a theory such as ZFC or PA. Clearly, to prove theorems about FOL we need assumptions such as the axiom of choice and those are theorem about this formalized FOL.


Positive integer multiples of an irrational mod 1 are dense I'm not sure how to solve this one.
Thank you!

$2.$ For any $\alpha\in \mathbb R$ we define $$\lfloor \alpha \rfloor = \max_{n\in\mathbb Z}\{\,n\mid n\leq \alpha\,\}$$ and $$\alpha\bmod 1 = \alpha - \lfloor \alpha \rfloor$$ Let $\alpha$ be irrational.
(a) Given $n\in\mathbb N$ show that $\{\,k\alpha\bmod1\mid k\in\mathbb N\,\}\cap\left[0,\frac{1}{n}\right]\neq\emptyset$
(b) Prove that $\{\,n\alpha\bmod 1\mid n\in\mathbb N\,\}$ is dense in $[0,1]$.

What I show below doesn't strictly follows the items, but still shows the numbers are dense.

STEP 1 Pick any $n\in\Bbb N$. Consider the $n+1$ distinct numbers $$x_k=k\alpha-\lfloor k\alpha\rfloor=\{k\alpha\}\; ;\;k=0,1,2,\ldots,n$$

These are $n+1$ numbers that fit into $n$ places, namely

$$\left[0,\frac 1 n\right),\ldots,\left[1-\frac{1}n,1\right)$$

By the Dirichlet's principle there must exist at least a pair of them which fall in the same interval of length $\dfrac 1n$.

STEP 2 We obtained $$\frac{k}{n} \leqslant \left\{ {{k_1}\alpha } \right\} < \left\{ {{k_2}\alpha } \right\} < \frac{{k + 1}}{n}$$

for some $k=0,\ldots,n$. We then know that $$0 < \left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\} \leqslant \frac{1}{n}$$

But note that $$\begin{align}
  \left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\} &= \left\{ {\left\{ {{k_2}\alpha } \right\} - \left\{ {{k_1}\alpha } \right\}} \right\} \cr
   &= \left\{ {{k_2}\alpha  - {k_1}\alpha  - \left( {\left\lfloor {{k_2}\alpha } \right\rfloor  - \left\lfloor {{k_1}\alpha } \right\rfloor } \right)} \right\} \cr
   &= \left\{ {{k_2}\alpha  - {k_1}\alpha  - {\text{integer}}} \right\} \cr
   &= \left\{ {\left( {{k_2} - {k_1}} \right)\alpha } \right\} \end{align} $$

Now, you may as well try and prove the following:

  Let $G$ be an additive subgroup of $\Bbb R$. Let $G^+$ denote the positive elements of $G$. Then

  $(1)$ If $\inf G^+=\alpha >0$, $G=\alpha\Bbb Z$

  $(2)$ If $\inf G^+=0$, $G$ is dense in $\Bbb R$.

Hint

For $(1)$. Show that $\alpha\in G$.

If not, pick $\epsilon =\alpha/2$ in the defintion of infimum, and  $g,g'\in G$ such that $\alpha  \leqslant g < g' <\alpha  + \frac{\alpha }{2}$. Look at $g'-g$.

   Then $g'-g\in G$ and $g'-g\leq \alpha/2<\alpha$ which is impossible.

Now pick $g>0$. Look at $g-\alpha \left\lfloor {\dfrac{g}{\alpha }} \right\rfloor$. Use the definition of integer part to show it must be zero.

   Thus $g = \alpha \left\lfloor {\dfrac{g}{\alpha }} \right\rfloor  \in \alpha {\Bbb Z}$. Since opposites are in $G$ too, $(1)$ is proven.

For $(2)$, pick any $x\in \Bbb R$. We know we can find $y>0$ in $G$ with $0<y<\epsilon$. Let $n=\left\lfloor {\dfrac{x}{y}} \right\rfloor $. What can you deduce from $$n \leqslant \frac{x}{y} < n + 1\text{ ? }$$

   We have $ny\in G$ by additivity, and $$yn \leqslant x < yn + y \Rightarrow 0 \leqslant x - yn < y < \varepsilon $$

Hint:
Check your definition of irrational. You can, by multiplying with $10^d \in \mathbb{N}$, generate an arbitrary small $[k\alpha]$.


$Re\left((1-z^2)^{\frac{1}{2}}+iz\right) > 0$ for $z\in\mathbb{C}\setminus\{z: Im(z) =0, |z|\ge1\}$ I need to prove that $$Re\left((1-z^2)^{\frac{1}{2}}+iz\right)>0,\quad Re\left(-(1-z^2)^{\frac{1}{2}}+iz\right)<0$$
if $z\in\mathbb{C}\setminus \{z: Im(z) =0, |z|>=1\}$.

The expontential of power $1/2$ means the principal square-root. The two inequalities can be verified easily when $Im(z)=0$ and $|z|<1$. I have no clue about how to handle the case where $Im(z)\ne 0$. I see that

$$\left((1-z^2)^{\frac{1}{2}}+iz\right)\left(-(1-z^2)^{\frac{1}{2}}+iz\right)=-1.$$

This means I just need to prove one of the inequalities and the other holds readily since the arguments of them differ by $\pi$.

Replace $z$ with $z=x+iy$ and simplify the expression. Eventually isolate the square root function and square both sides. Move all terms back to one side and take the real part of it.

The real part I get is $1-2y^2 > 0$ which when solved for gives $y^2 < \frac{1}{2} $. So my conclusion is that the real part of your expression is greater than zero for any values of $y^2$ that are less than $\frac{1}{2}$.


Definition of Global Convergence I am confused by the notion of "global convergence" as used in numerical optimization literature, and did not find an exact definition for that yet.
Now I try to double check my understanding here.

It clearly is NOT concerned with convergence to the global minimum.
Does it mean convergence to a local minimum regardless of the initial point? (How about local maximum, saddle point?) What methods do not have global convergence? Those that with selection of bad initial point can stop in an arbitrary non-critical point?

Also, a globally convergence method applied on a convex function, gives the global minima, right?

See the definition in http://papers.nips.cc/paper/3646-on-the-convergence-of-the-concave-convex-procedure.pdf.
The points generated by an algorithm with this property

converge for any initial point.
converge to a stationary point.

As far as I know optimisation would be the search for maxima/minima.

If you'd imagine your function being a mountain chain, the optimum would equal a peak
or a valley.

To find the minima or maxima you could calculate the gradient of the function
and starting from an initial point change the values of this initial point
into the direction the gradient is pointing in until you reach the peak/valley,
which would -- in most cases -- be a local optimum.

Global convergence  would hence be convergence to the global optimum regardless
of the initial starting point.

Hope this helps.

Edit:
Global Convergence is indeed the convergence to Minima/Maxima regardless
of the initial starting point.


Linear independence of columns and rows of a matrix. I read somewhere in my Algebra 1 notes something about how if a matrix has $n$ linearly independent rows (or it might have been columns) then it's rank is $n$. If this is the correct statement can someone please explain this, I can't seem to understand why this is true.
   However, if this statement has been quoted incorrectly, please can you correct it. Apologies if there is an error in my quotation, I can't find my notes.

I refer to rank as dim(Im(A)) i.e. the dimension of the A's image space.

The rank is the dimension of the column space (a.k.a. range or image) of a matrix. The column space is the span of the columns of the matrix. So one way to think of the rank is the size of a basis of the column space.

Some consequences:

If the matrix has $n$ columns, and $k$ of the columns are linearly independent, then the rank is at least $k$, since the span of these $k$ columns (which has dimension $k$) is a subspace of the column space.
If all $n$ columns of the matrix are linearly independent, then the columns themselves are a basis of the column space, so the rank is $n$.

The image of the matrix will be the span of the images of the basis vectors, i.e. the columns of the matrix. If there are $n$ linearly independent columns then the rank will evidently be $n$.

To see why the first line holds take a vector x=$\sum_{i=1}^n x_i \mathbf{e_i}$. First see that the image of a basis vector is the column of the matrix M since for the $k^{th}$ basis vector $(M\mathbf{e_k})_i =M_{ij}e_j=M_{ij}\delta_{jk}=M_{ik}$ which is the $i^{th}$ element of the $k^{th}$ column. Now we can apply $M$ to each component of x in the basis so $M\mathbf{x}$ would be in the span of the columns of M.

You can proceed similarly for the rows.