Fourier Analysis of Prime Counting Function I was thinking about the following:

Denote $\pi(x)$ as the prime counting function such that:
$$
\pi(x) = \#\text{ of prime numbers}\leq x
$$
It is well known from the prime number theorem that
$$
\pi(x) \sim \frac{x}{\ln x}
$$
and
$$
\pi(x) \sim \text{Li}(x),\quad \text{Li}(x)=\int_2^x\frac{1}{\ln t}\,dt.
$$
Note the following:
if $A(x) \sim B(x)\implies A(x)/B(x) = 1 \text{ as }x\to\infty$ and $A(x), B(x)\to\infty \text{ as }x\to\infty$.

If $C(x)$ is a function such that $n \leq C(x) \leq k$ then:

$$(A(x) + n)/B(x) \leq (A(x) + C(x))/B(x) \leq (A(x) + k)/B(x)$$

$$A(x)/B(x) + n/B(x) \leq A(x)/B(x) + C(x)/B(x) \leq A(x)/B(x) + k/B(x)$$

If x is taken to infinity:

$$1 + 0 \leq A(x)/B(x) + C(x)/B(x) \leq 1 + 0$$

$$\rightarrow A(x) + C(x) ~ B(x)$$

What interests me is that since we already know that:

$$\pi(x) \sim x/\ln(x)$$

and from above that $x/ln(x) +$ any number of functions of the form $A(b(x)) ~ \pi(x)$

Can we not try to do some sort of fourier analysis on the function:

$$\pi(x) - x/\ln(x) $$

or

$\pi(x) - \text{Li}(x)$?

The purpose of $\zeta(s)$ is to perform Fourier analysis on $\pi(e^u)$ :

$$\zeta(s) = \prod_p \frac{1}{1-p^{-s}} \quad \implies \quad \ln \zeta(s) = \sum_{p^k} \frac{p^{-sk}}{k}$$

i.e. it is the $\color{red}{\text{Laplace transform}}$ of $\sum_{p^k} \frac{1}{k}\delta(u-\ln p^k)$ $\qquad \scriptstyle \text{(or if you prefer, }\hat{f}(\xi ) =\ln \zeta(\sigma+2i \pi \xi) \text{ is the Fourier transform of } f(u) = \sum_{p^k} \frac{1}{\scriptstyle k \ p^{ \sigma k}}\delta(u-\ln p^k)$)

Integrating by parts, you get
$$\ln \zeta(s) = s \int_0^\infty J(e^u) e^{-su}du$$
where $J(x) = \sum_{p^k < x}\frac{1}{k} = \pi(x) + \mathcal{O}(\sqrt{x})$.

i.e.
$$\boxed{\frac{\ln \zeta(s) }{s} = h(s) + \int_0^\infty \pi(e^u) e^{-su}du}$$
where $h(s) = \int_0^\infty (J(e^u)-\pi(e^u))e^{-su}du$ is analytic and bounded for $Re(s) > 1/2+\epsilon$.

You can perform Fourier analysis on $\pi(x)$. Here are two plots of a Fourier series for $\pi'(x)$, i.e. the derivative of $\pi(x)$, with a frequency evaluation limit $f=1$ for the first plot and $f=4$ for the second plot. The two plots below illustrate the general relationship that this Fourier series always evaluates to $2f$ times the step size of $\pi(x)$ at integer values of $x$.

Fourier Series for π'(x) Evaluated at f=1

Fourier Series for π'(x) Evaluated at f=4


How to prove that $x^p \cdot \mathrm{e}^{-x} = O(x^{-2})$ When $x$ tends to $+\infty$ and where "O" is the Big-O-Notation

We recall the definition of big-O. That is $f(x) = O(g(x))$, if there exists constants $C,N$ such that $|f(x)| \leq C|g(x)| , \forall x > N$. For this problem, it suffices to consider cases.
Case 1: $p \leq 0$. Here it suffices to consider a constant $N  \geq 1$, for which $e^{-x} \leq x^{-2}$, $\forall x > N$, because $x^{p}e^{-x} \leq e^{-x} \leq x^{-2}$. But $e^{-x} \leq x^{-2}$, $\forall x > 2$. Hence, $x^{p} e^{-x} \leq x^{-2}$, $\forall x > 2$, when $p \leq 0$.
Case 2: $p > 0$. We want to find a constant $N$ for which $x^{p} e^{-x} \leq x^{p} x^{-(p+2)} = x^{-2}$. Then we can find an $N$ for which $e^{-x} \leq x^{-(p+2)}$, $\forall x > N$. However,
$$
e^{-x} \leq x^{-(p+2)} \iff \frac{x}{ln(x)} \geq (p+2)
$$
Choose $N = (p+2)^{4} \implies \frac{(p+2)^{4}}{4\cdot (p+2)ln(p+2)} \geq 1$, for $p > 0$. Altogether using the definition of big-O, we have that for arbitrary $p \in \mathbb{R}$, let $C = 1$. Then $x^{p}e^{-x} \leq x^{-2}$, $\forall x > N = \max\{(p+2)^{4}, 2\}$. Hence $x^{p}e^{-x} = O(x^{-2})$

We want to prove $x^pe^{-x}=O(x^{-2})\iff x^{p+2}e^{-x}=O(1)$

In other words we want to show that $|x^{p+2}e^{-x}|$ is bounded.

This is obviously true since $$\forall r\in\mathbb R,\lim\limits_{x\to+\infty}x^re^{-x}=0$$

Thus it is bounded for instance by $1$ provided $x\gg 1$ is sufficiently large.

To show that exponential kills any monomial, we can deduce it from $\dfrac{\ln(x)}x\to 0$ at infinity.

In particular it is bounded by $\frac 1{r+1}$ for $r>0$ and $x\gg 1$ sufficiently large.

Thus $(r+1)\ln(x)\le x\iff x^{r+1}\le e^x\iff x^re^{-x}\le \frac 1x\to 0$

The case $r\le 0$ does not pose an issue since in this case $x^r\le 1$ for $x\ge 1$ and $x^re^{-x}\le e^{-x}\to 0$ is easy.

The last point about $\ln(x)$ growth can be proven like this:

For $x\ge 1$ then $\frac 1x\le 1$ and we deduce that $f(x)=x-\ln(x)\nearrow$ since $f'(x)\ge 0$.

In particular $x-\ln(x)\ge f(1)=1\ge 0$ so $\ln(x)\le x$

If $x\ge 1$ so is $\sqrt{x}$ and we have $\ln(\sqrt{x})=\frac 12\ln(x)\le\sqrt{x}$

After dividing by $x$ we get $\dfrac{\ln(x)}x\le \dfrac 2{\sqrt{x}}\to 0$ at infinity.


How to prove that $\mathbb R$ with usual metric and $\mathbb R$ with the discrete metric are not homeomorphic. I know that two metric spaces are homeomorphic if there is a function from one to another such that $f$ is continuous, one-to-one, onto, with continuous inverse $f^{-1}$.
I know how to prove two metric spaces are homeomorphic.

I don't understand how to prove that two metric spaces are not homeomorphic.
I have to prove using knowledge I have of open sets, closed sets, limit and continuity.
I cannot use compactness, disconnectedness, etc.

Let $f\colon \mathbb{R}\to\mathbb{R}_d$ be a homeomorphism, where the domain has the usual metric and the codomain the discrete metric.

The sequence $(1/n)_{n>0}$ converges to $0$ in $\mathbb{R}$ with the usual metric, so $(f(1/n))_{n>0}$ should converge to $f(0)$ in $\mathbb{R}_d$. But in a discrete space only the eventually constant sequences converge.

Consider connectedness. Alternatively, one space is separable, and the other is not. Alternatively, the convergent sequences in one space are eventually constant, and you can count how many there are.


Find the radius of convergence of a series centered at a point a) Let $f(z) = \sum_{n≥0} z^n$. Find the radius of convergence of the series.
b) For all $t \epsilon \mathbb R$, $t<1$, find the radius of convergence of the Taylor series of $f$ centered at $t$. (Hint: let $w = z-t$ and write $f(z) = f(w+t)$)

My attempts:
a) The radius of convergence is $\frac{1}{\alpha}$, where $\alpha := lim_{n→\inf}$ sup n$\sqrt{|a_n|}$.
But here $|a_n| = 1$ and so the limit (as defined above) is equal to $1$.
Hence $R=1$
b)$f(z) = f(w+t)$ = $\sum_{n≥0} {(t+w)}^n$ = $\sum_{n≥0} \sum_{k=0}^n \binom{n} {k} t^{n-k} w^k$
But $\binom{n} {k} = \frac{n!}{k!(n-k)!}$
Let $a_n = \frac{n!}{k!(n-k)!}$
And so lim as $n$ tends to infinity of $\frac{a_{n+1}}{a_n} = 1$.
And hence $R=1$
Are my attemps correct? Any other solutions please? Thank you

Your answer for (a) is correct. For (b) I think is correct and here is another way, let $1 > t \in \mathbb{R}$.  Then the expansion of $f$ centered at $t$ is
$$ \sum_{n \geq 0} (z-t)^n $$
Which as you've already shown, converges for $|z-t|<1$.  Let $z=x+iy$, then $f$ converges on the set
$$ \{z : |z-t|<1 \} = \{ z : |z-t|^2<1 \} = \{ x+iy : (x-t)^2 + y^2 <1 \} $$
which is exactly the open disc of radius $1$ centered at $t$. So radius of convergence is $R=1$


equivalent regular expr Given a regular expression $r_1$, is $(r_1^*)^* = (r_1^*)$. I know that this is true but I do not know how to prove it.
Would it be adequate to set $r_1$ to a regular expression such as $(a+b)$ for example? Then showing that all the elements are found in both the RHS and LHS? i.e. $L((r_1^*)^*)$ is a subset of $L(r_1^*)$ and vice versa?
I tried to use an FA to show that they both would be accepted by the same FA but I don't think that is sufficient for a proof. for instance this
FA
Any sort of direction or help would be much appreciated.

The following result holds for any language $L$ (regular or not): $(L^*)^* = L^*$. Actually, a similar result holds for any monoid $M$. Given a subset $S$ of $M$, define $S^*$ as the submonoid generated by $S$, that is, the smallest submonoid of $M$ containing $S$. Equivalently $S^*$ is the smallest subset $N$ of $M$ containing $S$ and $1$ (the identity of $S$) and such that for all $x, y \in N$, $xy \in N$.

I claim that $(S^*)^* = S^*$. Indeed, since $(S^*)^*$ is a monoid containing $S$, it contains the smallest monoid containing $S$, namely $S^*$. Thus $S^* \subseteq (S^*)^*$. To prove the opposite inclusion, let $N = S^*$. Then $N$ contains $S^*$ (by definition) and $1$ and for all $x, y \in N$, $xy \in N$ (since $S^*$ is a monoid). Therefore $N$ contains the smallest monoid with these properties, namely $(S^*)^*$. Thus $S^* = (S^*)^*$.

It is straightforward that $(r_1*)$ is contained in $(r_1*)*$ .

Next, prove that $(r_1*)(r_1*)$ is contained in $(r_1*)$.  (Use double induction.)

Finally, prove that $(r_1*)*$ is contained in $(r_1*)$.  (Use induction.)


Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $f(mn)=f(m)+f(n)+3f(m)f(n)$ for $m,n \geqslant1.$ Prove that there exists no bijective function $f: \Bbb{N} \to \Bbb{N}$ such that $$f(mn)=f(m)+f(n)+3f(m)f(n)$$ for $m,n \geqslant1.$

This was a problem from a Putnam practice book and I couldn't seem to figure out how to show this. Initially, I started to wonder that if the function were bijective then it has to be injective as well as surjective. So trying to determine if its injective let $a,b \in \mathbb{N}$ then if it were injective $$f(mn)=f(ab) \implies f(m)+f(n)+3f(m)f(n) =f(a)+f(b)+3f(a)f(b)$$ but this doesn't seem to help since I don't have explicitly $f$. How should I look at this problem?

Let's assume $0\in\mathbb{N}$, since otherwise the condition $m,n\ge1$ would be redundant. Then, $m=n=1$ implies $f(1)=0$. If we define $g(n)=3\,f(n)+1$, we obtain $$g(mn)=g(m)g(n).\tag{mult}$$ So we have a completely multiplicative bijection between $\mathbb{N}$ and $\mathbb{M}=\{m\in\mathbb{N}: m=1\bmod3\}$. Now the numbers $10,34,55,187$ all belong to $\mathbb{M}$, so there must be natural numbers $n_1, n_2, n_3, n_4$ with $$g(n_1)=10,\quad g(n_2)=34,\quad g(n_3)=55,\quad g(n_4)=187.$$ The $n_i$ must be different, and they must be primes, because $d|n_i$ implies $g(d)|g(n_i)$ through (mult), and $g(n_i)$ by definition don't have non-trivial divisors in $\mathbb{M}$. But then, we have
$$g(n_1n_4)=g(n_1)\,g(n_4)=g(n_2)\,g(n_3)=g(n_2n_3),$$ and since $g$ is injective, $$n_1n_4=n_2n_3,$$ but this is impossible for four different primes.

1.Show that the function is well defined
2.Prove that there exist a kernel.
3. Prove that identity maps to element of the kernel
If you can't prove 1 and 2 the function or mapping is not bijective.
In other words, you can prove that the function is both injective and surjective( i.e if the inverse of the function maps back to every element in the domain) as required.


Finding area in a bounded region How would I solve the following problem?

  Sketch the region bounded by the graphs of $f(x)=x^2-4x$ and $-3x^2$ and then find the area of the region.

I have made a sketch for this question but what I am not sure how to do is find the area.

I have found the intersections $x=0$ and $x=1$

$x^2-4x=-3x^2$

$4x^2-4x=0$

$x=0,x=1$

$$f(x)=\int_0^1 x^2-4x-(-3x^2)(dx)$$

$$f(x)=\int_0^1 4x^2-4x(dx)$$

$$G(X)=\frac{4x^3}{3}-2x^2$$

So I did

$$A=\frac{4}{3}(1)-2(1)^2-0$$

$$A=\frac{-2}{3}$$

But how can an Area be negative?

Note that on the interval $[0, 1]$, $-3x^2 > (x^2 - 4x)$.  So the "upper curve" we are interested in is $-3x^2$ and the lower curve $(x^2 - 4x)$.

Always subtract the curve that is UNDER, from the curve that is OVER.

So you need for your integrand to be $$\int_0^1 -3x^2 - (x^2 - 4x) \,dx$$

Otherwise, your work was fine! You found your bounds correctly, and you integrated and evaluated correctly, just wrong choice of the order of subtraction.

Determine the  point of Intersect bounded by the  curve f(x) = x to the  power 4 and 2x to the power 2 and  g(x)= 2x to the power 2, (( i=(2) , (j) = 3,((( -3 is equal and less than x < 3)))


How can one efficiently generate n small relatively prime integers? The definition of small is that they have O(lg n) bits. One way is just to test the integers 2,3,... for primality and keep the first n primes, but this takes at least O(n log n) time (times the cost of primality testing) using a naive algorithm. Is it possible to do it in linear time?

You can find the first n primes in time $n\log n/\log\log n$ with the Atkin-Bernstein sieve.  This is better than your naive $n\log^5n$ or so algorithm (or drop the exponent by 1 using pretesting with M-R), but still superlinear.

I assume S need not be prime (just relatively prime)

For S= {2, 15}

Goldbach method : N = 2 + M

                = 2 + 30

                = 32 (not relatively prime with 2)


Why is the conjugate function closed? I have the fenchel-dual function $f^*: \mathbb {R}^n\rightarrow \mathbb {R}\cup \infty : f^*(y)=\sup_{x\in A}x^ty-f (x) $ with $A \subset \mathbb {R} $ being closed. Why is this function closed?

$f^*$ is closed because for every $x\in A$, the function
$$ y\mapsto x^ty-f(x)$$
is closed, and thus is $f^*$ as a supremum of closed functions.


Let $f(x)$ be a differential real function defined on the real line. If $f(0)=0$ and $f'(x)=[f(x)]^2$, then $f(x)=0$ foi any $x$. Again, $f:\mathbb{R}\to\mathbb{R}$ is differentiable, $f(0)=0$, and $f'(x)=[f(x)]^2$ for every $x$. A friend suggested the following argument:

If exists $c$ such that $f(c)\neq0$, there exists an interval $I$ around $c$ such that $f(x)\neq0$ if $x\in I$ (because $f$ is continuous since it is differentiable). In that interval, we could define $g(x)=x+\frac{1}{f(x)}$. This function $g$ would be differentiable and $g'(x)=0$. Then $g(x)$ is constant, for example, $k$. Then, $f(x)=\frac{1}{k-x}$ for $x\in I$

But I don't know where to find an absurd. What should I do next?

I think I should use the fundamental theorem of calculus and try to find an absurd with $f(x)=\int_0^x f'(t)dt=\int_0^x [f(t)]^2 dt$, but I also didn't got anywhere.

Thanks in advance.

Suppose $f(x)$ exists on some closed interval $[0,\delta]$.  Let $F=F(\delta)$ be its maximum value on this interval.  If $x\in(0,\delta)$, then your integral equation $f=\int f^2$ yields
$$
|f(x)|\le \int_0^xf(t)^2dt\le x F^2.
$$
Hence,
$$
F(\delta)\le \delta F(\delta)^2.
$$
If $F(\delta)\neq 0$, then
$$
1/\delta\le F(\delta).
$$
This contradicts continuity of $f(x)$ at $x=0$, since we can choose $\delta$ arbitrarily small.

Consider the differential equation (y=f(x)) $y'=y^2$.  The solution is $y=\frac{1}{c-x}$ where c is an arbitrary constant.  If f(0)=0, then c must be infinite, so that f(x) is identically 0.


example of a linear map in real space for which $0$ is the only eigenvalue but the map is not nilpotent? What would be an example of a linear map in real space for which $0$ is the only eigenvalue but the map is not nilpotent? Only hint is enough.

If

$$A=\left(\begin{matrix}
0&1&0\\
-1&0&0\\
0&0&0
\end{matrix}\right)$$
then $\chi_A=-x(x^2+1)$ (which we calculate easily using block matrices) so $0$ is the only eigenvalue in $\mathbb R$.
The matrix $A$ isn't nilpotent since $A^3=-A$.

This cannot happen in finite dimension.

Hint: Such a map would have a Jordan normal form with zeros on the diagonal.


Taylor series of $f(x^2)$ If you know the taylor series for $f(x)$ can you find the taylor series for $f(x^2)$ by letting $x = x^2$? The taylor series in question is $\cos(x^2)$

I know the taylor series for $\cos(x)$ is $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{2n}$. Is the taylor series for $\cos(x^2)$ just $\displaystyle \sum_{n=0}^\infty \frac{(-1)^n}{(2n)!} x^{4n}$? If this works, does it work for all functions $f(x)$ with a closed form solution for the taylor series?

Indeed you can, so long as $x^2$ is within the region of convergence for the taylor series.

It really is just like any other function. If $f(x)=\sum_n a_nx^n$, then $f(y)=\sum_n a_ny^n$ and $f(x^2)=\sum_n a_nx^{2n}$.

You can calculate this directly from the Taylor series definition. Which is given here: http://en.wikipedia.org/wiki/Taylor_series (sorry I can't figure out how to type out LateX formulae here) You basically calculate the nth derivate and divide by n! at a. Normally Taylor series calculated at 0 are called MacLaurean series.
I don't think you can directly plug it into the Taylor series for cos(x) because the derivative of cos(X^2) is different.


Generalization of Cauchy Residue theorem to Multi-dimensional holomorphic functions We know Cauchy Residue theorem from the Complex analysis. however I wonder if there is a kind of Generalization of Cauchy integral and Residue theorem to the complex multidimensional holomorphic function too ?
if such theorem exists, please mention any possible application of this theorem too.
Can this theorem help in Iterative elimination of roots of a polynomial equations system ?

Cauchy's integral formula has a straight-forward generalization to polydiscs. If $f$ is holomorphic on (a neighbourhood) of the polydisc $\Omega = \mathbb{D}(z_1,r_1) \times \mathbb{D}(z_2,r_2) \times \cdots \times \mathbb{D}(z_n,r_n)$, and $a \in \Omega$, then

$$
f(a) = \frac{1}{(2\pi i)^n} \int_{|z_1|=r_1} \int_{|z_2|=r_2}\cdots\int_{|z_n|=r_n}
\frac{f(z_1, z_2, \ldots, z_n)}{(z_1-a_1)(z_2-a_2)\cdots(z_n-a_n)}\,dz_1\cdots dz_n.
$$

The really fascinating thing is that we are integrating just on a tiny part of the boundary of $\Omega$.

For more general domains $\Omega \subset \mathbb{C}^n$, you have the Bochner-Martinelli integral formula, and there are lots of other integral representation formulas. Keywords: Leray and Koppelman integral formulas. See for example Range: Holomorphic Functions and Integral Representations in Several Complex Variables for details.

Regarding generalizations of residues to several complex variables, see this recent question.

A function of a single complex variable generates harmonic functions in the X-Y plane.  This leads to an interesting way of thinking about Cauchy's integral formula and Green's integral formula for harmonic functions.

Cauchy's formula:  A boundary line integral of the complex function can determine the value of the function within the enclosed area.
Green's formula: A boundary surface integral of a harmonic function can determine the value of the function within the enclosed volume:


Which CSL rules hold in Łukasiewicz's 3-valued logic? CSL is classical logic. So I'm talking about the basic introduction and elimination rules (conditional, biconditional, disjunction, conjunction and negation).

I'm not talking about his infinite-valued logical theory, but the 3-valued one where any atomic sentence can be given T,F or I. (See the Wikipedia article here:

http://en.wikipedia.org/wiki/Three-valued_logic

try it out by 3 valued truthtables :)

out of my head all  except the conjunction introduction , negation elimination and negation introduction rule are valid .
but i can be wrong

added after Comment by Spoonwood

The problem with the conditional introduction rule is that it is normally a discharging rule and in Lukasiewicz  3 valued logic

(P -> (P-> Q)) -> (P -> Q) (contraction) is not a tautology in Lukasiewicz  3 valued logic :
(I use i instead of 1/2, the  designated value is 1, * is the problematic line)

 P | Q || (P -> (P -> Q)) -> (P -> Q)
---|---||-----------------|-----------
 1 | 1 ||    1     1      1     1
 1 | i ||    i     i      1     i
 1 | 0 ||    0     0      1     0
 i | 1 ||    1     1      1     1
 i | i ||    1     1      1     1
 i | 0 ||    1     i      i*    i
 0 | 1 ||    1     1      1     1
 0 | i ||    1     1      1     1
 0 | 0 ||    1     1      1     1

so you should just not be able to deduce it
but using the standard conditional introduction rule
you can:

1 ||___ P -> (P -> Q)               Assumption
2 |||__ P                           Assumption
3 |||   P -> Q                      1,2 Conditional elimination
4 |||   Q                           2,3 Conditional elimination
5 ||<   P -> Q                      2-4 Conditional introduction
6 |<    (P -> (P -> Q)) -> (P -> Q) 1-5 Conditional introduction

so the conditional introduction rule as used here is not a valid rule in Lukasiewicz 3 valued logic.

You could have a rule

if P then Q -> P (without discharging an assumption)
but that is not the standard rule.

While writing this I was also wondering about the or elimination rule, it depends a bit which or elimination rule you use, some are valid some are not, the same applies to the Not introduction rule.

In general non-discharging rules are valid, discharging rules are not

1) I believe that Conditional introduction works:
     My experience is that the problem lies in getting a getting valid derivation from a set of premises, given the other rules that don't work.

2) Conditional elimination does not.
   This is the chief thing that cripples Lukasiewicz logic as a logic.
   Modus Ponens and the transitivity of the conditional both fail  for the Lukasiewicz conditional, as they properly should, given that it allows conditionals with a value of I. If your conditionals are doubtful, you should not expect inference using them to be valid.
    (It is not generally known that this can be remedied: Define Spq as NCCpqNCpq and use that for the conditional)

3) Biconditional introduction works.
4) Biconditional elimination works.
5) Conjunction introduction works.
6) Conjunction elimination works
7) Disjunction introduction works.

8) Disjunction elimination does not work.
9) Negation introduction does not work.
10) Negation elimination does not work.
    Classically, 9 and 10 depend on the Law of the Excluded Middle, which doesn't hold in Lukasiewicz logic.


Notation in regards to limits Do we define infinity as a "limit"? Or do we simply say that the limit doesn't exist as the function/series diverges?

I'm calculating the limit of a function, turns out to be infinity, but I am not sure whether to just write "infinity" after an equal sign, or to say "this doesn't exist".

Also, can I even use the term: "Lim f(x) = " without knowing whether the limit exists beforehand?

you can say the limit is positively (or negatively) divergent and it would be correct, but among mathematicians no one looks bad on:
$$\lim_{x\to x_0}f(x)=+\infty$$
because the result of a limit is not necessarely in $\mathbb{R}$, most analists  use this definition: $\mathbb{\hat R}=\mathbb{R}\  \cup \{-\infty,+\infty\}$ (the extended real line) and if you assume that the result lies in $\mathbb{\hat R}$ then it's perfectly rigorous to write the limit result using the notation above. (for more info look: here)

$\infty$ is not a number.

Consider for example if limit of some function tends to infinity at some point c

than it means that if you get closer and closer to that point than the function just doesn't have bounds.

We say that the limit exists but it is not bounded which is indicated as $\infty$.

i.e.,

if you get closer and closer towards c then your limit doesn't have any bounds.


Prove that $\vdash p \lor \lnot p$ is true using natural deduction I'm trying to prove that $p \lor \lnot p$ is true using natural deduction. I want to do this without using any premises.

As it's done in a second using a truth table and because it is so intuitive, I would think that this proof shouldn't be too difficult, but I not able to construct one.

First assume any thesis $\alpha$ as an assumption.  Then show that the thesis can have the same scope as ApNp.  Then infer C$\alpha$ApNp.  Then prove $\alpha$ and use conditional-out.  I use Polish notation, and I can give you formation rules if you like:

Cpp assumption

NApNp assumption

p assumption

ApNp 3 A-in

⊥ 3, 4. K-in

Np 3-5 N-in

ApNp 6 A-in

⊥ 2, 7 K-in

ApNp 2-8 N-out

CCppApNp 1-9 C-in

p assumption

Cpp 11-11 C-in
ApNp 10, 12 C-out

Note that we did not prove |-ApNp at any point (though we did prove Cpp|-ApNp).


Positive Integer Divison Proof Prove that $6|n(n + 1)(n + 2)$ for any integer $n ≥ 1$

I have attempted to do this but never seem to prove it.

As $3$ positive consecutive integers, either $n$, $n+1$, or $n+2$ must be divisible by $3$.

As $2$ positive consecutive integers, either $n$ or $n+1$ must be divisible by $2$.

Therefore, since $n(n + 1)(n + 2)$ is divisible by $2$ and $3$ ($2$ and $3$ are relatively prime numbers), it must be divisible by $6$.

If $n$ is even: $n = 2k \implies n(n+1)(n+2) = 2k(2k+1)(2k+2) = 2k(4k^2+6k+2)= 8k^3+12k^2+4k = 6k^3+12k^2 + 6k+2k^3-2k$. Observe that if $3 \nmid k\implies k^2-k = 0\pmod 3$ by Fermat Little's theorem. Thus $2k^3 - 2k = 0\pmod 6$. Thus $6 \mid n(n+1)(n+2)$. If $n$ is odd, then $n = 2k+1\implies n(n+1)(n+2) = (2k+1)(2k+2)(2k+3)= (m-1)m(m+1)$ with $2 \mid m = 2k+2$. So if $3 \nmid m\implies m^2-m = 0\pmod 3$ again by Fermat Little's theorem. Thus $6 \mid n(n+1)(n+2), \forall n \ge 1$.


Why does an argument similiar to 0.999...=1 show 999...=-1? I accept that two numbers can have the same supremum depending on how you generate a decimal representation. So $2.4999\ldots = 2.5$ etc.
Can anyone point me to resources that would explain what the below argument that shows $999\ldots = -1$ is about?
Here is the most usual proof I see that $0.999\ldots = 1$:

$x=0.999\ldots$
$10x=9.999\ldots$
$10x - x = 9$
$x=1$

Using this same argument template I can show $999\ldots=-1$:

$x= \ldots9999.0 $
$0.1x= \ldots9999.9$
$0.1x - x = 0.9$
$x=-1$

What might this mean?
Edit from one of the comments:
$$\sum_{k=0}^{\infty}{9 \cdot 10^k}=-1$$

If you want to understand the mathematics behind these things, it is all based upon the notions of 'convergence' and of 'limits'. If you read any first course in analysis textbook you will find the concept rigorously treated there.

Basically this is the point: Whenever you write 0.999... you are writing down a numeral that represents the 'limit' obtained when an infinite summation $\frac{9}{10}+\frac{9}{100}+\frac{9}{1000}+...$ is performed. Since we can prove that this sum 'converges' to some real number (namely 1), we are justified in treating the numeral 0.999... as representing some real number.

However, whenever you write down 999... I presume you are writing a numeral to represent the limit obtained when an infinite summation $9+90+900+...$ is performed. Since this limit does not converge to any real number, (it 'diverges'), we are not justified in treating the numeral 999... as any real number. So it does not make sense to divide it by ten, or take it away from itself.

We usually denote such divergent limits by the numeral $\infty$, but this does not denote a real number, and there is no consistent way to define operations such as $\infty - \frac{1}{10}\infty$.

I hope this helps and I hope you are motivated to think more about these things :)

Folks have put forth the $p$-adic interpretation, but another way of understanding this is to consider it in the context of divergent series. One of the ideas behind the study of divergent series is to view $\sum$ as a linear functional from a particular domain $C(\mathbb{R})$ to $R$, where $C(\mathbb{R})$ is the vector space consisting of all real sequences with convergent series. This operator has many useful properties in and of itself, evidenced by how much we care about series in general.

However, we may also pose the question, Can we extend $\sum$ to an operator, call it $S$, on a vector space $V \subseteq \mathbb{R}^{\mathbb{N}}$ containing $C(\mathbb{R})$? In a very trivial way, the answer is yes, we can. To see this, recall that the axiom of choice implies that every vector space has a basis, so let $\{ e_{i} \in C(\mathbb{R}) : i \in I \} \subseteq \{ e_{j} \in V : j \in J \}$ be bases for $C(\mathbb{R}), V$ respectively. Then we can set
$$ S (e_{j}) = \begin{cases}
\sum (e_{j}) & e_{j} \in C(\mathbb{R}) , \\
0 & e_{j} \in V \setminus C(\mathbb{R}) .
\end{cases} $$
However, this is not very satisfying. Linearity is not the only thing we like about summation. For example, consider the shift operator $\sigma : (a_{k})_{k \in \mathbb{N}} \mapsto (a_{k + 1})_{k \in \mathbb{N}}$. A property of $\sum$ is that $\sum ((a_{k})_{k \in \mathbb{N}}) = a_{1} + \sum(\sigma((a_{k})_{k \in \mathbb{N}}))$. We like this, for example, because it means that when doing convergence tests we know that no matter how $(a_{k})_{k \in \mathbb{N}}$ "starts", if it eventually "looks like" a sequence whose sum converges, then we still have convergence; that is, changing any finite number of terms in a sequence won't change whether or not its series converges. I will call this the Front-End Property (FEP).

It is important to note that the proof you listed in the OP that $0. \overline{9} = 1$, though very common, is as written incomplete. In particular, it assumes (rightfully, of course) that $0. \overline{9}$ is defined. You can show this, for instance, by appealing to the fact the sequence $(0.9, 0.99, 0.999, \ldots )$ is monotonic and bounded by $1$, and thus convergent. From this, we consider that $0. \overline{9} \equiv \sum (9 \cdot 10^{-k})_{k \in \mathbb{N}}$, i.e. $0. \overline{9}$ is defined as the summation of the sequence $(0.9, 0.09, 0.009, 0.0009, \ldots)$. We then use the front-end property. Let $x = \sum (9 \cdot 10^{-k})_{k \in \mathbb{N}}$. Then
\begin{align*}
10 x & = \sum(10 \cdot (0.9 \cdot 10^{-k}))_{k \in \mathbb{N}} \\
& = (10 * 0.9) + \sum(10 \cdot (0.9 \cdot 10^{-(k + 1)}))_{k \in \mathbb{N}} \\
& = 9 + \sum(0.9 \cdot 10^{-k})_{k \in \mathbb{N}} \\
& = 9 + x \\
\Rightarrow 9x & = 9 \\
\Rightarrow x = 1 .
\end{align*}
But line one to line two was just the FEP! What you actually proved in the OP is that $S(10 \cdot (0.9 \cdot 10^{-k}))_{k \in \mathbb{N}} = 1$ for any operator $S$ with the FEP. You're mimicking the same argument to show that $\overline{9} = -1$, i.e. you're showing that IF $S$ has the FEP, THEN $S(9, 90, 900, 9000, \ldots) = - 1$ for a linear map $S : \mathbb{R}^{\mathbb{N}} \to \mathbb{R}$. It just happens that $\sum$ is not defined at $(9 \cdots 10^{k})_{k \in \mathbb{N}}$.

So in essence we're saying, "So we know we have this thing called summation that we can do to sequences that does neat stuff, and its defined in terms of these other sequences of partial sums converging. But we can also talk about this summation as a linear transformation, which is just the same as saying I can multiply by a scalar or add the sequences and get what I expect. But summation has other properties as an operator than linearity. We mentioned the FEP, but it's also what we may call positive, i.e. if $a_{k} \geq 0$ for all $k$, then $\sum (a_{k})_{k \in \mathbb{N}} \geq 0$. So we may ask, Can we extend this to all of $\mathbb{R}^{\mathbb{N}}$ and keep FEP and positivity? As you've shown, no.

Note also that if you apply your standard geometric series formula for when the ratio is $10$, you'll get $\overline{9} = - 1$.


Matrix representation of the Quaternions? Can anyone explain how why the matrix representation of the quaternions using real matrices is constructed as such?

I am not sure exactly what you mean by asking "... why the ...". "Why" questions can be hard to answer satisfactory in math.

The claim is that the Quaternions $\mathbb{H}$ are isomorphic (as $\mathbb{R}$-algebras) to the given set of matrices. The isomorphism looks like this:

$$
\phi: a + bi + cj + dk \longmapsto \begin{pmatrix}a & b & c & d  \\  -b & a & -d & c \\ -c &d &a& - b\\ -d& -c & b& a\end{pmatrix}.
$$
To "understand" why this is true, you "simply" check that this is an isomorphism.

You check for example that $\phi$ is bijective, which is clear from the construction.

Then you check that $\phi$ is an algebra homomorphism, so you need for $x,y\in \mathbb{H}$ and $\lambda \in \mathbb{R}$:

$\phi(xy) = \phi(x)\phi(y)$ for $x,y\in\mathbb{H}$
$\phi(x+y) = \phi(x) + \phi(y)$
$\phi(\lambda x) = \lambda\phi(x)$

The last two are not difficult to check. The first one requires a bit of work.

Even though this does not answer the minus signs are where there are in the matrix, I highly recommend that you try to prove that $\phi$ is a homomorphism. This exercise will make you more familiar with the Quaternions.

But note if you check property $3$ above you would need (as a special case)
$$
\phi((bi)(bi))  = \phi(ib)\phi(ib).
$$
That is you would need
$$
\begin{pmatrix} -b^2 & 0 & 0& 0  \\  0 & -b^2 & 0 & 0 \\ 0 &0 &-b^2& 0\\ 0& 0 & 0& -b^2\end{pmatrix} = \begin{pmatrix} 0 & b & 0& 0  \\  -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}\begin{pmatrix} 0 & b & 0& 0  \\  -b & 0 & 0 & 0 \\ 0 &0 &0& -b\\ 0& 0 & b& 0\end{pmatrix}.
$$
So here you can see that you "need" the minus on all the $b$'s. In this case it comes down to the fact that $i^2 = -1$.

If It is not a misunderstanding, the question of "why as such (possible others?)" is equivalent to the problem of finding out ALL 4x4 real matrix representations of the Quaternions. The fact is: all those matrix representations are conjugated to each other. In other words, this is the only 4x4 real matrix representation of the Quaternions up to equivalent.


$x-12\%=100$, find $x$ I need to pay someone after deducting $12\%$ commission. I want to pay him round amount of $10,000$ What is the formula to establish the $112\%$ figure so that once I have deducted my comm I will be left with $10,00$.
I know the answer is not $11200$ because $11200 - 12\% = 9856$.

Please help me with a formula that will calculate $x - 12\%= 10,000$
or $x - 12\% = 7000$  or $x-12\% = 5000$.

Thanks so much,
Maurice

One may observe that, in general,
$$
x - a\%\:x=b
$$ is just
$$
\left(1-\frac{a}{100} \right)\times x=b
$$ giving, if $a \neq 100$,

  $$
x=\frac{b}{\left(1-a/100\right)}
$$

see its quite simple sorry for misinterpreting first time. so here we assume total amount to be $x$ thus you want $12\%$ commision of $x$ and extra $10K$. so our equation becomes $\frac{12x}{100}+10000=x$ thus on solving you get $x=11363.63..$ now i hope i have interpreted your question correctly.


$\sup_nX_nA)<\infty$ Let $(X_n)_{n\in\mathbb N}$ be a sequence of independent random variables. Show that $\sup_nX_n<\infty$ almost surely iff there exists $A>0$ such that, $\sum_nP(X_n>A)<\infty$

By Borel-Cantelli we have

$$\sum_nP(X_n>A)=\infty\Longleftrightarrow P(\limsup_nX_n>A)=1$$

Hence $\forall A>0$;

$\sum_nP(X_n>A)=\infty\Rightarrow 1=P(\limsup_nX_n>A)<P(\sup_nX_n>A)$

Can you give a hint for the other direction, Thanks.

I'll assume the random variables are real-valued.

'direction 1'

Let $A_n^c := \{X_n > A\}$. By BCL1, we have

$$P(\limsup A_n^C) = 0$$

$$\to P(\liminf A_n) = 1$$

$$\to P(\lim A_n) = 1$$

$$\to \lim P(A_n) = 1$$

$$\to P(\bigcap_{n=1}^{\infty} A_n) = 1$$

$$\to \prod_{n=1}^{\infty} P(A_n) = 1 \ \text{Why?}$$

$$\to \forall n \in \mathbb N, P(A_n) = 1$$

$$\to \forall n \in \mathbb N, P\{X_n \le A\} = 1$$

$$\to P( \sup_{n \ge 1} (X_n) < \infty) = 1$$

'direction 2'

Show the contrapositive:

By BCL2, we have

$$P(\limsup A_n^C) = 1 \ (\forall A > 0)$$

$$\to P(\liminf A_n) = 0$$

$$\to \lim_{m \to \infty} P(\bigcap_{n=m}^{\infty} A_n) = 0$$

$$\to \lim_{m \to \infty} \prod_{n=m}^{\infty} P(A_n) = 0$$

$$\to \sup_{m \ge 1} \prod_{n=m}^{\infty} P(A_n) = 0$$

$\to \forall m \ge 1, \exists n \ge m$ s.t. $P(A_n) = 0$

$$\to P(X_n \le A) = 0 \tag{*}$$

Now suppose on the contrary that $P[\sup X_n < \infty] = 1$.

$$\to \sup[X_1, X_2, ...] < \infty \ \text{a.s.}$$

$\to \exists N \ge 0$ s.t.

$$P(\bigcap_{n=1}^{\infty} X_n \le N ) = 1$$

$$\to \prod_{n=1}^{\infty} P(X_n \le N ) = 1$$

$$\to P(X_n \le N ) = 1 \forall n \ge 1$$

Choose $A = N$ in $(*)$. ↯

$$\therefore, \forall A > 0, \sum_{n=1}^{\infty} P(X_n > A) =\infty \to P[\sup X_n < \infty] < 1$$

QED


How is it, that $\sqrt{x^2}$ is not $ x$, but $|x|$? As far as I see, $\sqrt{x^2}$ is not $x$, but $|x|$, meaning the "absolute". I totally get this, because $x^2$ is positive, if $x$ is negative, so $\sqrt{y}$, whether $y = 10^2$ or $y = -10^2$: $y$ is positive.

But then I remember that $\sqrt(x)$ is the same as $x^{1/2}$ and thus, $\sqrt{x^2}$ is the same as $x^{1}$.

So, as far as I get it, $\sqrt{x^2} = x^{2/2}$.

But then, I can cancel $\frac{2}{2}$ to $1$. So: $\sqrt{x^2} = x^1 = x$.

Where does the absolute value in this derivation of the calculation come from?

I get it why its there: Because $(-x)^2 = x^2$, so the radicand is always positive. But from $\sqrt{x^n} = x^{n/2}$ I dont understand how it gets there.

This is why you need to pay attention to the hypotheses of a theorem! The exponent laws you've learned are almost surely only for integer exponents or for positive bases: until you have learned multi-valued functions and complex exponentiation, you should always be suspicious whenever you're both using non-integers in exponents and negative numbers in bases in the same problem. (presumably, you'd understand the nuances once you've fully learned how to deal with complex analysis)

$√x$ = $(x)^{1/2}$ represents only the principle value of square root of x. Its the exponential representation of principle square root.

Principle square root: The non-negative (positive) square root of a number is called its Principal Square Root.

Thus, $√(x^2)$ = $(x^2)^{1/2}$ = x is the principle square root (positive)

While, $√(x^2)$ = $±x$ ; both positive and negaive square roots.


Using the Dirac delta function for taking limits In trying to calculate:

$$\displaystyle \lim\limits_{a\to0} \frac{2x}{a^3} e^{-\frac{x^2}{a^2}}. \tag{1}$$

I initially tried to make an argument with the taylor series (looking at the limit for each term individually), but there were some edge cases that were difficult to deal with( when x=0) and the whole argument was somewhat ugly. I've been advised some properties of the dirac-delta function (and it's derivative) can bee applied to this problem, but I don't quite see how. The function itself does not behave like a dirac-delta, nor is any convenient piece of it (so far as I can see).

How would using the function be relevant to these types of problems?

Well, there is a so-called heat kernel representation of a nascent delta function
$$\delta(x)~=~\lim_{a\to 0^+} \delta_a(x) , \qquad \delta_a(x)~:=~\frac{1}{\sqrt{\pi}a}e^{-\frac{x^2}{a^2}}, $$
$$\delta^{\prime}(x)~=~\lim_{a\to 0^+} \delta^{\prime}_a(x) , \qquad \delta^{\prime}_a(x)~=~-\frac{2x}{\sqrt{\pi}a^3}e^{-\frac{x^2}{a^2}}. $$

In other words. OP's formula (1) is a representation of the derivative of the Dirac delta distribution (times a factor minus square root of pi).

Using Dirac's delta function here would be an overkill. From what you wrote, we see that $a$ is the variable of this limit and $x$ is a parameter. Simply consider two cases for the parameter $x$:

If $x=0$, then $\displaystyle \lim\limits_{a\to0} \frac{2x}{a^3} e^{-\frac{x^2}{a^2}}=\displaystyle \lim\limits_{a\to0} 0=0$.
If $x\neq0$, then $\displaystyle \lim\limits_{a\to0} \frac{2x}{a^3} e^{-\frac{x^2}{a^2}}=2x\lim\limits_{a\to0} \frac{e^{-\frac{x^2}{a^2}}}{a^3}$, and then complete it using L'Hôpital's rule or Taylor series, or just recall than exponential functions grow or decay much faster than power functions.


Limits - a different approach $\lim_{x \to\infty }(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$.

How do I calculate the value of l and m?

My try: I know questions having limit tending to infinity can be solved by dividing the numerator and denominator by greatest power of $x$.But it got me nowhere in this question. Any help appreciated.

$$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5}{x^2+2x+3}+lx+m)=10$$

$$\lim\limits_{x \to \infty}(\frac{x^3+4x^2+3x+5+lx^3+2lx^2+3lx+mx^2+2mx+3m}{x^2+2x+3})=10$$

$$\lim\limits_{x \to \infty}(\frac{(1+l)x^3+(4+2l+m)x^2+(3+l+2m)x+(5+3m)}{x^2+2x+3})=10$$

Note that as the limit exists, the coefficient of $x^3$ has to be $0$, so we get $l=-1$

And also the ratio of the leading coefficient ($x^2$)$=10$, so we get $4+2l+m=10$ and thus $m=8$

$\dfrac{x^3+4x^2+3x+5}{x^2+2x+3} \approx_{\infty} x\implies \lim_{x \to \infty} (x+lx+m) = 10\implies l = -1, m = 10.$


Different Representation of Linear System Currently working on my Linear Algebra homework and I have come across several problems of the same form that I haven't the slightest clue how to approach.  Keep in mind I haven't taken math classes for some time now, so I am a bit rusty.  I have searched online, and in my book, to no avail.

I certainly don't want anyone to do the problem for me, just (1) what problems of these forms are called - e.g. so I can search on google for info on how to solve it; and (2) maybe some links for where I can get more information.

Sorry for such a stupid post, but I am running short on time, and was hoping someone might be able to help.

One of the problems I'm referring to:

In this case here you have the equations of two planes (in coordinate form). Now you have to get the equation of the intersecting line (parameter form).

Note that you can write the equations in the form
$A \begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = \begin{bmatrix} 9 \\ 8 \end{bmatrix} -\begin{bmatrix} 5 \\ -3 \end{bmatrix} x_3$, so we get
$\begin{bmatrix} x_1 \\ x_2 \end{bmatrix} = A^{-1}\begin{bmatrix} 9 \\ 8 \end{bmatrix} -A^{-1}\begin{bmatrix} 5 \\ -3 \end{bmatrix} x_3$.

In the question you can take $x_3 = s$.


What does "does not cancel on the interval" mean? I have a question which asks me to consider the system
$$f(x)=a+bx+\frac{P(x)}{Q(x)},\qquad x\in I,$$
where $P$ and $Q$ are polynomials, and stipulates "$Q(x)$ does not cancel on the interval $I$ ".
What does this mean?
Does this mean $Q(x)\neq0$ on this interval, or that $Q(x)\neq P(x)$ on this interval, or perhaps something else? I have tried Googling the expression but couldn't find anything, and I haven't came across it before.

It might mean that whenever $Q(x)=0$, then $P(x)\ne 0$. And the other way around, $P(x)=0\implies Q(x)=0$, but that is probably the less interesting case.
As $P,Q$ are polynomials, the usual use of "cancel" is that a common factor can be removed, here that could be interpreted the way that $P,Q$ do not have a common factor that has roots inside that interval.
But it might also be an unfortunate word choice and the condition to express was that "$Q$ does not vanish on $I$", which indeed then means that there are no roots of $Q$ in $I$.

It means $Q(x) \neq 0$ for all $x\in I$.


Upper bound on Huffman codeword length I am reading Elements of Information Theory by Cover and Thomas and have been unable to find an upper bound on the length of a codeword in a Huffman code, either in this book or on the web.  Does one exist?  If so, could someone provide an outline of a proof and an example achieving this bound?  Please assume that we don't know the number of symbols to be encoded in advance, only the probability of a given codeword.

The length of a codeword is simply the length of the path from the root of the tree in the construction of the Huffman code to the leaf corresponding to the symbol that it codes for. If you know the total number of symbols to be coded for, then you know the number of leaves, so what is the maximum path length possible?

After edit

If you only have the probability of a codeword, then you need to have a finer grasp of the Huffman tree. You should of course first understand the proof of the Huffman code optimality, since the core property is the key to most of its other properties. Recall that in the tree the weight of a leaf is just the probability of the symbol and the weight of an internal node is just the sum of the children's weights.

The core property is that in any optimal tree, for any two nodes $a,b$, if $a$ is heavier than $b$ then $a$ is not below $b$, otherwise swapping the subtrees will give a better tree. This is key to proving that any optimal tree can be transformed into another optimal tree that agrees with the code generated by the Huffman algorithm.

This same core property is key for this question too. We want for any node an upper bound on its depth given a lower bound on its weight. Equivalently, we want an upper bound on its weight given a lower bound on its depth. But if we go from the root to the lightest child at each step, its weight must decrease by a factor of at least $2$ at each step. Thus the lightest node at depth $k$ has weight at most $2^{-k}$. But any node at depth $k+1$ is at least as light as the lightest node at depth $k$ by the core property, and hence we have an upper bound on its weight. From this it is easy to get an upper bound on the depth given the weight.

I would have posted a reply to the previous answer instead but I do not have enough reputation so I will try to make this answer worth while still. The previous response by user21280 is not percise. A lightest path to a node at depth $d$ does not always exist. For instance take the distribution $(1/3, 1/3, 1/3)$, the lightest child of the root has weight of $1/3$ but has no children, thus there is no lightest path of depth 2. It is possible to observe that this example is scalable and this indeed is no fluke, there is an underlying truth which is: huffman trees, although minimize the average depth of a node and are optimal in that sense, do not necessarily achieve the $\text{ideal length} + 1$ bound for all nodes. In order to prove the $\text{entropy} + 1$ bound one can consider the Shannon Fano code which, whereas not necessarily optimal, achieves the $\text{ideal length} + \text{at most $1$}$ for each node. This means that SF codes achieve the $\text{entropy} + 1$ bound, and through optimality of the huffman code on the average depth one can deduce that the previous upper bound holds for huffman trees as their average depth is upper bounded by that of SF trees.

Proving that there are codes that achieve ideal length up to 1 (without using SF codes) can be done simply using Kraft's inequality (Kraft-McMillan) as can be seen at this link on pages 10-13.


Create unique number from 2 numbers is there some way to create unique number from 2 positive integer numbers? Result must be unique even for these pairs: 2 and 30, 1 and 15, 4 and 60. In general, if I take 2 random numbers result must be unique(or with very high probability unique)

Thanks a lot

EDIT: calculation is for computer program,so computational complexity is important

The sort of function you are looking for is often called a pairing function.

A standard example is the Cantor pairing function $\mathbb{N}\times\mathbb{N}\rightarrow\mathbb{N}$, given by:
$\pi(a,b) = \frac{1}{2}(a+b)(a+b+1) + b$.

You can find more information here: http://en.wikipedia.org/wiki/pairing_function

Apologies for resurrecting this ancient question, but I've noticed that there are collisions in the results of the Cantor pairing function.

For example, I've noticed that when a and b are 0.0 and 0.0, or 0.6 and 0.0 the result is the same: 0.48.

Is this function expected to work for non-discrete numbers, or does this have something to do with the fact they the numbers are 0.0 and 0.0 in the former case?


Exponential relationship issue I read this relation and I am not sure why this is true, is it I can't see why it would be?

$$(e^{ -i\pi/2})^{ -ix}\approx ie^{-\pi x/2} $$ I get that $e^{i\pi/2}=-i$, but I can't see why this relation would be true.

You have to be careful working with complex exponentials because of problems with multivalued logarithms.  The exponential $a^b$ for $a,b \in \mathbb{C}$  is usually defined in terms of the multivalued logarithm as $e^{b \log a}$.

As a multivalued expression, we have
\begin{align*}
(e^{-i \pi/2 })^{-ix} &= e^{-ix \log (e^{-i \pi/2 })} \\
&= e^{-ix (-i\pi/2 + 2\pi i k) } && (k \text{ an integer; } k = 0 \text{ for principle log})\\
&= e^{-x\pi/2 + 2\pi x k} \\
&= e^{2\pi k x}e^{-x\pi/2}
\end{align*}

Therefore, for your statement to be true (in the sense that for one branch it works) we would have to have
\begin{align*}
e^{2\pi k x}e^{-x\pi/2} &= i e^{-x\pi/2} \\
\iff e^{2\pi k x} &= i \\
\iff e^{2\pi k x} &= e^{i \pi / 2} \\
\iff 2 \pi k x &\equiv i \pi / 2 \pmod {2\pi i}
\end{align*}

For this to be true, $x$ must be purely imaginary.  Let $x = i(m + r)$ where $m$ is an integer and $0 \le r < 1$ real.
\begin{align*}
2 \pi k (i(m + r)) &\equiv i \pi / 2 \pmod {2\pi i} \\
\iff 2 \pi i m k + 2 \pi i k r &\equiv i \pi / 2 \pmod {2\pi i} \\
\iff mk + kr &\equiv 1 / 4 \pmod {1} \\
\iff kr &\equiv 1 / 4 \pmod {1}
\end{align*}

In summary: the only time your statement could actually be sort of true is when $x$ is purely imaginary, and the fractional part of the imaginary part of $x$ (i.e. $r$ above) multiplied by some integer $k$ has a fractional part of $1/4$.

Use basic index laws.

$$
(e^a)^b = e^{ab}
$$
What do you get when you do this?


How to show field extension equality I've seen similar field extension questions on SE, but nothing with a third root, and I'm having trouble adapting any of those solutions to this problem.

So I'm trying to prove that $\mathbb{Q}(\sqrt{2}+{5}^{1/3})=\mathbb{Q}(\sqrt{2},{5}^{1/3})$.

Now, $\mathbb{Q}(\sqrt{2},{5}^{1/3})$ contains both $\sqrt{2}$ and ${5}^{1/3}$, so $\mathbb{Q}(\sqrt{2}+{5}^{1/3})\subseteq\mathbb{Q}(\sqrt{2},{5}^{1/3})$, right?

How then do I go about proving the reverse, that $\mathbb{Q}(\sqrt{2},{5}^{1/3})\subseteq\mathbb{Q}(\sqrt{2}+{5}^{1/3})$? I not even sure where to start. Any advice, tips?

An idea

$$r=\sqrt2+\sqrt[3]5\implies r^3-3\sqrt2\,r^2+6r-2\sqrt2=5\implies$$

$$(r^3+6r-5)^2=\left[(3r^2+2)\sqrt2\right]^2$$

The above shows that $\;\dim_{\Bbb Q}\Bbb Q(\sqrt 2+\sqrt[3]5)= 6\;$ (why? Show the polynomial $\;(r^3+6r-5)^2-\left[(3r^2+2)\sqrt2\right]^2\in\Bbb Q[x]\;$ is irreducible ), but since $\;\dim_{\Bbb Q}\Bbb Q(\sqrt2\,,\,\sqrt[3]5)=6\;$ (why?) and we already know $\;\Bbb Q(\sqrt2+\sqrt[3]5)\le\Bbb Q(\sqrt2\,,\,\sqrt[3]5)\;$ we get equality.

Actually I found an even simpler way, after receiving some hints from my professor.

Let $\alpha=\sqrt{2} + \sqrt[3]{5}$, and notice that $(\alpha^2-2)/2\alpha=\sqrt[3]{5}$. Thus both $\sqrt{2}$ and $\sqrt[3]{5}$ are in $Q(\sqrt{2}+\sqrt[3]{5})$, and the two field extensions must be equal.


Final step of exercise 11.7 from Atiyah-Macdonald ($\dim A[x]=\dim A+1$) Ex. 11.7 from Atiyah-Macdonald is basically to prove $\dim A[x]=\dim A+1$ for $A$ noetherian.

From exercise 11.6, we get $\dim A[x]\geq\dim A+1$, so we are left to prove "$\leq$".

I've followed the hint and proved that $\text{ht}(p[x])=\text{ht}(p)$ for every $p\subset A$ prime, but I'm still not sure why this implies $\dim A[x]\leq\dim A+1$.

Here is what I've thought so far:

if $P_1\subsetneq...\subsetneq P_k$ is a maximal chain in $A[x]$, then $P_k$ is maximal and $k=\dim A[x]$. Suppose that $m:=P_k\cap A$ is a maximal ideal of $A$. That way, $A/m$ is a field and $A[x]/m[x]\simeq A/m[x]$ is a P.I.D. Since $P_k$ is maximal, $\overline{P_k}\in A[x]/m[x]$ is maximal, so $\text{ht}(\overline{P_k})=1$, which means $P_k$ is minimal prime of $m[x]$. That way, $k=\text{ht}(P_k)=\text{ht}(m[x])+1=\text{ht}(m)+1\leq\dim A+1$.

The problem with what I did is that it depends on $P_k\cap A$ being maximal. Is this necessarily the case? I feel like this is crucial, otherwise the hint wouldn't work. How do I prove that?

If it isn't true, how can I conclude that $\dim A[x]\leq\dim A+1$?

We'll use the following fact which occurs in the hint to Exercise 11.6 p. 126 of the book:

($\star$) If $f:A\to A[x]$ is the natural embedding, then the fiber of
$$
f^*:\operatorname{Spec}(A[x])\to \operatorname{Spec}(A)
$$
over a prime ideal $\mathfrak p$ of $A$ is order isomorphic to the spectrum of
$$
k\otimes_AA[x]\simeq k[x],
$$
where $k$ is the residue field at $\mathfrak p$.

We denote the Krull dimension of the ring $A$ by $\dim A$, and the height of the prime ideal $\mathfrak p$ by $h(\mathfrak p)$.

Let's prove the inequality $\dim A[x]\le1+\dim A$.

It suffices to prove
$$
h(\mathfrak p)\le1+h(\mathfrak p^{\operatorname c})
$$
for all prime ideal $\mathfrak p$ of $A[x]$ by induction on $h(\mathfrak p^{\operatorname c})$.

If $h(\mathfrak p^{\operatorname c})=0$, the statement follows from ($\star$).

Assume $h(\mathfrak p^{\operatorname c})>0$. It suffices to show

(a) $h(\mathfrak q)\le h(\mathfrak p^{\operatorname c})$ for all prime ideal $\mathfrak q$ of $A[x]$ strictly smaller than $\mathfrak p$.

Indeed, there is a chain
$$
\mathfrak q_0\underset\ne\subset\cdots\underset\ne\subset\mathfrak q_{n-2}\underset\ne\subset\mathfrak q\underset\ne\subset\mathfrak p
$$
with $n=h(\mathfrak p)$, and we get $h(\mathfrak q)\ge n-1=h(\mathfrak p)-1$, that is, $h(\mathfrak p)\le1+h(\mathfrak q)$.

Let $\mathfrak q$ be a prime ideal as in (a).

If $\mathfrak q^{\operatorname c}\underset\ne\subset\mathfrak p^{\operatorname c}$, we have $h(\mathfrak q^{\operatorname c})<h(\mathfrak p^{\operatorname c})$, and the induction hypothesis implies $h(\mathfrak q)\le 1+h(\mathfrak q^{\operatorname c})\le h(\mathfrak p^{\operatorname c})$.

If $\mathfrak q^{\operatorname c}=\mathfrak p^{\operatorname c}$, we have $\mathfrak q=\mathfrak p^{\operatorname c}[x]$ by ($\star$). This implies $h(\mathfrak q)=h(\mathfrak p^{\operatorname c})$.

Consider the quotient map $q : A[x] \rightarrow A[x]/(x) \cong A$. Clearly $q(P_k) = P_k \cap A$, and if $q(P_k)$ not maximal then by Zorn's lemma $q(P_k) \subset Q$ where $Q$ is a maximal ideal of $A$. However, this implies $ P_k \subset q^{-1}(Q)$ where $q^{-1}(Q)$ is a prime ideal, a contradiction.


Generalized angle sum identity for $\arctan$? The angle sum  identity for arctan is:

$$\arctan (\alpha)+\arctan(\beta)=\arctan\left(\frac{\alpha+\beta}{1-\alpha\beta}\right)$$

I was wondering if there exists a relationship for any linear combination of $\arctan$ functions:

$$A\arctan (\alpha)+B\arctan(\beta)$$

$A>0, B>0$ and both real numbers.

Edit: Yes, such a formula indeed exists:
https://functions.wolfram.com/ElementaryFunctions/ArcTan/16/03/03/
Old post
I know this question has been asked a long time ago, but I figured this would still be of use. There must be such a formula for a linear combination of $\arctan$s! I don't know what exactly it looks like, but I accidentally came across this which is similar (long, painful way to get there at all, but anyway):
These two expressions (maybe someone else can format more nicely, sorry:) Mathematica code at the end)
$$A:=\frac{1}{2 a^{3/2}}
\\\left(\frac{ \left(2 \sqrt{a} \sqrt{a+b}-2 a+b\right) \sqrt{(b-a) \left(2 \sqrt{a} \sqrt{a+b}+2 a+b\right)} \arctan\left(\frac{b \left(\left(\sqrt{a}-\sqrt{a+b}\right) \tan \left(\frac{\phi}{2}\right)+\sqrt{a}\right)}{\left(\sqrt{a}-\sqrt{a+b}\right) \sqrt{(b-a) \left(2 \sqrt{a} \sqrt{a+b}+2 a+b\right)}}\right)}{b \left(\sqrt{a+b}+3 \sqrt{a}\right)}\\
-\frac{ \left(b-2 \left(\sqrt{a} \sqrt{a+b}+a\right)\right) \sqrt{(b-a) \left(-2 \sqrt{a} \sqrt{a+b}+2 a+b\right)} \arctan\left(\frac{b \left(\left(\sqrt{a+b}+\sqrt{a}\right) \tan \left(\frac{\phi}{2}\right)+\sqrt{a}\right)}{\left(\sqrt{a+b}+\sqrt{a}\right) \sqrt{(b-a) \left(-2 \sqrt{a} \sqrt{a+b}+2 a+b\right)}}\right)}{b \left(\sqrt{a+b}-3 \sqrt{a}\right)}\right)$$
$$B:=-\frac{\sqrt{b-a} \arctan\left(\sqrt{\frac{a}{b-a}} (\sin (\phi)+\cos (\phi))\right)}{2 a^{3/2}}$$
don't look very similar. However, deriving them with Mathematica by $\phi$ yields the same expression
$$\frac{(a-b) (\cos (\phi)-\sin (\phi))}{2 a (a \sin (2 \phi)+b)}.$$
Consequently, I think that $A$ and $B$ differ by a constant, if at all.
What strikes me is that both the coefficients and the arguments to the $\arctan$s seem to contain complex conjugate-like terms everywhere, for example $\sqrt{a+b}-3 \sqrt{a}$, $\sqrt{a+b}+3 \sqrt{a}$ in the denominators. Maybe someone else can resolve the mystery and come up with a linear combination formula for $\arctan$s?
These two links are also quite insightful on such a linear combination :
1, 2
Mathematica code:
FullSimplify[D[((2 (-2 a+b+2 Sqrt[a] Sqrt[a+b]) Sqrt[(-a+b) (2 a+b+2 Sqrt[a] Sqrt[a+b])] ArcTan[(b (Sqrt[a]+(Sqrt[a]-Sqrt[a+b]) Tan[phi/2]))/((Sqrt[a]-Sqrt[a+b]) Sqrt[(-a+b) (2 a+b+2 Sqrt[a] Sqrt[a+b])])])/(b (3 Sqrt[a]+Sqrt[a+b]))-(2 Sqrt[(-a+b) (2 a+b-2 Sqrt[a] Sqrt[a+b])] (b-2 (a+Sqrt[a] Sqrt[a+b])) ArcTan[(b (Sqrt[a]+(Sqrt[a]+Sqrt[a+b]) Tan[phi/2]))/((Sqrt[a]+Sqrt[a+b]) Sqrt[(-a+b) (2 a+b-2 Sqrt[a] Sqrt[a+b])])])/(b (-3 Sqrt[a]+Sqrt[a+b])))/(4 a^(3/2)),phi],{a+b>0,b-a>0}]

FullSimplify[D[-((Sqrt[-a+b] ArcTan[Sqrt[a/(-a+b)] (Cos[phi]+Sin[phi])])/(2 a^(3/2))),phi],{a+b>0,b-a>0}]

note that ; this relation is always true, see below
$$arctan(\frac{1}{2})+arctan(\frac{1}{3})=arctan(\frac{\frac{1}{2}+\frac{1}{3}}{1-\frac{1}{2}\frac{1}{3}})\\arctan1=\frac{\pi}{4} $$but $$ arctan(2)+arctan(3)\neq arctan(\frac{2+3}{1-2.3})=-\frac{\pi}{4}$$it is because $$arctan(\frac{1}{2}),arctan(\frac{1}{2})<\frac{\pi}{2}$$but $$arctan(2),arctan(3)>\frac{\pi}{2}$$so sum of two term can not be $$-\frac{\pi}{4}$$


in the triangle ABC on the AC side, points M and N are chosen such that ABM = MBN = NBC in the triangle ABC on the AC side, points M and N are chosen such that <ABM = <MBN = <NBC It turned out that NB = BC. On the side AB, a point K was marked such that BK = BM.  Prove that AK + NC> AM.
I tried to get a triangle with sides AK NC and AM.
But I couldn't. So I don't know how to prove this

Geometric hint:
Consider the figure:
Following cases can be considered:
a-If BC=BF then triangle BCF is isosceles then NC=DE and  FD+DE>FE. This is a particular case.
b-If BC<AB then we have:
AK+KG=AK+DE>AM
Since $\angle AKM>90^o$ and:
$NC>DE>\frac{KM}{2}$
Therefore:
AK+NC>AM

let $BC=a$ and $\measuredangle ABC=3\beta$.
Thus, $$\measuredangle ACB=90^{\circ}-\frac{\beta}{2},$$
$$\measuredangle ANC=90^{\circ}+\frac{\beta}{2},$$
$$\measuredangle AMB=90^{\circ}+\frac{3\beta}{2}$$ and
$$\measuredangle BAC=180^{\circ}-\left(90^{\circ}+\frac{5\beta}{2}\right),$$ which says
$$90^{\circ}+\frac{5\beta}{2}<180^{\circ},$$ which gives $$0^{\circ}<\beta<36^{\circ}.$$
Now,  $$NC=2a\sin\frac{\beta}{2}$$ and by law of sines we can show that:
$$AM=\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ and
$$AK=AB-BK=\frac{a\cos\frac{\beta}{2}}{\cos\frac{5\beta}{2}}-\frac{a\cos\frac{\beta}{2}}{\cos\frac{3\beta}{2}}=\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}.$$
Id est, we need to prove that:
$$\frac{a\sin\beta\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+2a\sin\frac{\beta}{2}>\frac{a\cos\frac{\beta}{2}\sin\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or
$$\frac{\cos\frac{\beta}{2}\sin2\beta}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}+1>\frac{\cos^2\frac{\beta}{2}}{\cos\frac{5\beta}{2}\cos\frac{3\beta}{2}}$$ or $$2\cos\frac{\beta}{2}\sin2\beta+\cos4\beta+\cos\beta>1+\cos\beta$$ or
$$\cos\frac{\beta}{2}\sin2\beta>2\sin^22\beta$$ or
$$1>4\sin\frac{\beta}{2}\cos\beta$$ or
$$1>4\sin\frac{\beta}{2}\left(1-2\sin^2\frac{\beta}{2}\right)$$ or
$$1-2\sin\frac{\beta}{2}>2\sin\frac{\beta}{2}\left(1-4\sin^2\frac{\beta}{2}\right)$$
$$1>2\sin\frac{\beta}{2}\left(1+2\sin\frac{\beta}{2}\right)$$ or
$$\sin\frac{\beta}{2}<\frac{\sqrt5-1}{4},$$ which we got by looking for  domain of $\beta$.


Mathematical Name for Physical Gauge Symmetries In physics, when talking about a gauge transformation, we always mean two combined transformations. For example, a $U(1)$ gauge transformation is a combination of

$$ \psi \rightarrow e^{ia(x)} \psi $$

plus a simultaneous transformation on a different function

$$ A_\mu \rightarrow  A_\mu + i  \partial_\mu a(x) $$

Is there some mathematical term for this kind of combined transformation. Or is is something of the form

$$ U(1) \oplus R  \quad  ?$$

Let me first write this a bit differently, you are probably familiar with this.
If we consider gauge transformations $U(x)$ with an arbitrary gauge group $G$, the fields transforms as follows:
$$ \psi \mapsto U \psi \;, \quad A_\mu \mapsto U A_\mu U^{-1} + i (\partial_\mu U) U^{-1} \;. $$
(For $G = U(1)$ if you insert $U = e^{ia}$, you get your formulae back.)

Global Transformations:
Let us first pretend that the second term in the transformation of $A$ was not there, i.e. $A_\mu \mapsto U A_\mu U^{-1}$.
This is what happens for global gauge transformations where $U$ is constant.

Our notation hides the fact that $\psi$ actually transforms in some representation $R$ of $G$:
$\psi$ lives in some vector space $V$ and $U$ is an element of $G$, so when we write $U\psi$ we implicitly mean that $U$ acts on $\psi$ in that representation $R$.
(We do this because usually $G = SU(n)$ and $V = \mathbb C^n$ where there is a natural representation.)

The fields $A_\mu$ take values in the Lie algebra $\mathfrak g$ of $G$ and $G$ acts in the adjoint representation $Ad$ defined by $Ad(g)(A) = g A g^{-1}$.
What I want to show with all of this is:
You were on the right track.
Because the two fields $\psi$ and $A$ transform simultaneously, they transform in the direct sum of the two representations:
$$ R \oplus Ad $$

Gauge Transformations:
Let's consider now the complete picture:
The transformation behavior of $A$ is quite peculiar because the action $A_\mu \mapsto U A_\mu U^{-1} + i (\partial_\mu U) U^{-1}$ is not linear and therefore not a representation at all.
It is better to talk about this in completely different terms (literature recommendation: this link, and also chapters 9 and 10 of Nakahara).

A gauge transformation is actually a diffeomorphism of the principal $G$-bundle $P$.
The fermion field is a section of the vector bundle associated with $P$ over the representation $R$.
A gauge transformation acts naturally on the associated vector bundle, this can be described by a function $U$ like above in a local trivialization.

A connection on the principal bundle induces a connection on the associated vector bundle.
Locally, this connection is described by the vector potential $A$, which is a $\mathfrak g$-valued 1-form.
The connection on $P$ has a very natural transformation behavior under gauge transformation -- the transformation of the vector potential of $A$ results from this.


Randomly selecting a natural number In the answer to these questions:

Probability of picking a random natural number,
Given two randomly chosen natural numbers, what is the probability that the second is greater than the first?

it is stated that one cannot pick a natural number randomly.

However, in this question:

What is the probability of randomly selecting $ n $ natural numbers, all pairwise coprime?

it is assumed that we can pick $n$ natural numbers randomly.
A description is given in the last question as to how these numbers are randomly selected, to which there seems to be no objection (although the accepted answer is given by one of the people explaining that one cannot pick a random number in the first question).

I know one can't pick a natural number randomly, so how come there doesn't seem to be a problem with randomly picking a number in the last question?

NB: I am happy with some sort of measure-theoretic answer, hence the probability-theory tag, but I think for accessibility to other people a more basic description would be preferable.

It really depends on what you mean by the "probability of randomly selecting n natural numbers with property $P$". While you cannot pick random natural number, you can speak of uniform distribution.

For the last problem, the probability is calculated, and is to be understood as the limit when $N \to \infty$ from the  "probability of randomly selecting n natural numbers from $1$ to $N$, all pairwise coprime".

Note that in this sense, the second problem also has an answer. And some of this type of probabilities can be connected via dynamical systems to an ergodic measure and an ergodic theorem.

Added The example provided by James Fennell is good to understand the last paragraph above.

Consider ${\mathbb Z}_2 = {\mathbb Z}/2{\mathbb Z}$, and the action of ${\mathbb Z}$ on ${\mathbb Z}_2$ defined by

$$m+ ( n \mod 2)=(n+m) \mod 2$$

Then, there exists an unique ergodic measure on ${\mathbb Z}_2$, namely $P(0 \mod 2)= P(1 \mod 2)= \frac{1}{2}$.

This is really what we intuitively understand by "half of the integers are even".

Now, the ergodic theory yields (and is something which can be easily proven directly in this case)

$$\lim_{N} \frac{\text{amount of even natural numbers} \leq N}{N} = P( 0 \mod 2) =\frac{1}{2} \,.$$

There is no uniform distribution on the natural numbers.


The problem lies in two places: the first one in $$\frac{\partial Q}{\partial x} = 2ye^{y^2},$$ but this is only algebraic, even though it made your life much harder.

The second one is much more sensible. Look at Green's Theorem
$$\oint_C (P dx + Q dy) = \iint_D \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\,dA,$$ where D is CLOSED domain and $C = \partial D$.
Therefore, the integral you are suppose to have is

$$\int_C F \cdot dr = -\left[\iint_{\triangle} \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\, dA + \int_3^0 F(t,0) \cdot (1,0) \, dt\right],$$
where, $\triangle$ is the triangular domain of the question, $((0,0),(1,2),(3,0))$, and the the last term is the integral to bound you domain. Note that we integrate from $3 \to 0$ in order to close the domain, and we need the negative of RHS, because the chosen parametrization, that needs to respect the right-hand rule.

So, let
$$G(x,y) = \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = 2$$

We have
\begin{align}
\int_C F \cdot dr &= -\left[ \int_0^1\int_0^{2x} G(x,y) \, dy\,dx + \int_1^3\int_0^{-x+3}G(x,y) \,dy\,dx + \int_3^0 F(t,0) \cdot (1,0) \, dt,\right]\\
&= -\left[ \int_0^1\int_0^{2x} 2 \, dA + \int_1^3\int_0^{-x+3} 2 \,dA + \int_3^0 1 \, dt,\right]\\
&= -\left[ 2\int_0^1\int_0^{2x}  \, dA + 2\int_1^3\int_0^{-x+3}  \,dA - \int_0^3  \, dt,\right]\\
&=-\left[2\dfrac{1\times2}{2} + 2\dfrac{2 \times 2}{2} - 3 \right]\\
&=-3
\end{align}

The partial derivative of $Q$ regarding $x$ lacks a factor $2$.


Is the image of a curve under the projection morphism of a fiber product again a curve? If $X$ is a variety defined over the rationals and $f:X_\mathbb{C}\to X$ is the projection map of the fibered product, then is $f(C)$ is a curve in $X$? (Where $C$ is a curve in $X_\mathbb{C}$.)

No. Let $X=\Bbb A^2_{\Bbb Q}$ and let $C=V(x-\pi)\subset X_{\Bbb C}$. The preimage of the ideal $(x-\pi)\subset \Bbb C[x,y]$ in $\Bbb Q[x,y]$ is the zero ideal since $\pi$ is transcendental over $\Bbb Q$. Thus $f(C)$ contains a point with residue field of transcendence degree $2$ over $\Bbb Q$, so no matter what subscheme structure you put on it, it cannot be a curve over $\Bbb Q$.
Let me point out that what you're doing is a bit strange - in general, the question "what subscheme structure do I put on $f(Z)$ for a subscheme $Z\subset X$ and a map $f:X\to Y$" doesn't really have a good answer and may not even be answerable at all.


Weak solution of SDE Show that the SDE

$dX_t=dB_t-\frac{Bt}{1-t}dt$

has a weak solution on $[0,t]$ for $0\ge t<1$.

I tried it with Girsanov's theorem. But I can't prove that
$E[\exp(\int_0^t\left(\frac{Bs}{1-s}\right)^2ds]<\infty$ which would imply that $Z_t:=\exp(M_t-\frac{1}{2}\langle M\rangle_t)$ is a continuous martingale $\iff E[Z_t]=1 \quad \forall t\ge0.$ So I can't apply Girsanov.

Hope that it help you :to prove this $$E[\exp(\int_0^t\left(\frac{Bs}{1-s}\right)^2ds]<\infty$$
note that
    $0<s<t<1 \\$ so  $ \exists n_0: s^{n_0}<\epsilon$
$$\dfrac{B_s}{1-s}=B_s(1+s+s^2+s^3+...)\\\to \\B_s(1+s+s^2+s^3+...)<B_s .M$$


Show that if $p>5$ is prime, then $240 \mid (p^4-1)$ Show that if $p>5$ is prime, then $240 \mid (p^4-1)$.

By Fermat's Little Theorem $p^4\equiv 1\pmod 5$ and $p^4\equiv \left( p^2\right)^2 \equiv 1^2\equiv 1\pmod 3$. Also $p^4-1=(p-1)(p+1)\left(p^2+1\right)$. Exactly one of $p-1, p+1$ is divisible by $4$, so $(p-1)(p+1)\left(p^2+1\right)$ is divisible by $4\cdot 2\cdot 2$.

$$240=3\cdot5\cdot16$$

Now using Carmichael function $\lambda(240)=\cdots=$lcm$(2,4,4)=4$

So, if $(p,30)=1,p^4\equiv1\pmod{240}$

$p$ does not need to prime


Prove that Binet's formula gives an integer, using the binomial theorem I am given $$
F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}
$$
where, $\varphi = \frac{1 + \sqrt{5}}{2}$ and $\psi = \frac{1 - \sqrt{5}}{2}$.

The textbook states that it's equal to the $n$-th Fibonacci number $F_n$.
it is stated that since the Fibonacci numbers are integers, the number in $$
F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}
$$
is an integer as well. can you guys clarify this please?

  How do I go on proving that the number
  $$F_n = \frac{\varphi^n - \psi^n}{\sqrt{5}}$$
   is an integer using Newton's Binomial Theorem?

How do I go on proving that the number in is an interger using Newton's Binomial Theorem?

$$\newcommand{\b}[1]{\left(#1\right)}
\newcommand{\d}{{\rm d}}
\newcommand{\f}{\frac}
\newcommand{\s}{\sqrt}
\newcommand{\t}{\text}
\newcommand{\u}{\underbrace}\bf Answer$$

$$S=\f1{\s5}\b{\b{\f{1+\s5}2}^n-\b{\f{1-\s5}2}^n}=\f1{2^n\s5}\b{\sum_{k=0}^n\binom5k5^{k/2}-\sum_{k=0}^n(-1)^k\binom5k5^{k/2}
}=\f1{2^n\s5}\sum_{k=0}^n[1-(-1)^k]\binom5k5^{k/2}$$

For when k is even the terms cancel so we get only odd terms let wherin we let $k=2t-1;k\in\{1,3,5,..n\},t\in\{1,2,..(n+1)/2\}$
$$S=\f1{2^{n-1}\s5}\sum_{t=1}^{(n+1)/2}\binom5{2t-1}5^{t-1/2}=\f1{2^{n-1}}\sum_{t=1}^{(n+1)/2}\binom5{2t-1}5^{t-1}$$
Now it's not irrational anymore and you may use induction.

I think this question was asked before.

First, using induction prove,

  $$\Large{\phi^n = F_{n-1} + F_n \phi} \; \; \& \; \; \Large{\psi^n = F_{n-1} + F_n \psi}$$

Where $\phi = \frac{1 + \sqrt{5}}{2}$, and $\psi = 1 - \phi$.

Therefore,

  $$\Large{\phi^n - \psi^n = F_{n-1} + F_n \phi -  F_{n-1} - F_n \psi} \\ \Downarrow \\ \Large{\phi^n - \psi^n = F_ n ( \phi - \psi ) = F_n \sqrt{5}} \\ \Downarrow \\ \Large{\frac{1}{\sqrt{5}} ( \phi^n - \psi^n ) = F_n }$$

$F_n $ is integer, therefore $\frac{1}{\sqrt{5}} ( \phi^n - \psi^n ) \; $ is also integer.


Pigeonhole principle to prove that given $4$ numbers the difference between two numbers is divisble by $3$ "Use the Pigeonhole Principle to show that among any four numbers one can find two numbers so that their difference is divisible by $3$."

I am struggling with this supposedly basic question on one of our past paper as its only worth $3$ marks.

Hint : A number can be of the form $3k , 3k+1$ or $3k+2$. Since there are $4$ numbers , at least two of them leave the same remainder when divided by $3.$

What remainder would their difference leave when divided by $3 ?$

There are 3 possible remainders when we divide a num by 3.thus by pigeon hole principal we have 4 num.two of them must have same remainder when we divisible by 3.so we can write these n1 And n2 as
$N1=3q1+r,$
$N2=3q2+r,$
$N1-N2=3q1+r-3q2-r =3(q1-q2)$
Which is divisible by 3


Difference between Norm and Distance I'm now studying metric space. Here, I don't understand why definitions of distance and norm in euclidean space are repectively given in my book.

I understand the difference between two concepts when i'm working on non-euclidean space, but is there any even slight difference between these two concepts when it is $\mathbb{R}^k$?

All norms can be used to create a distance function as in $d(x,y) = \|x-y\|$, but not all distance functions have a corresponding norm, even in $\mathbb{R}^k$. For example, a trivial distance that has no equivalent norm is $d(x,x) = 0$, and $d(x,y) = 1$, when $x\neq y$. Another distance on $\mathbb{R}$ that has no equivalent norm is $d(x,y) = | \arctan x - \arctan y|$.

However, in general, when working in $\mathbb{R}^k$ the distance used is one induced by a norm, and 'unusual' distances are typically used to illustrate other mathematical concepts (eg, the $\arctan$ distance gives an example of an incomplete metric space).

The metric space/distance metric is a mathematical construction built up on topological space (definition of convergence, continuity, completeness, separability, connectedness, compactness etc. etc.). The distance operates with point sets and doesn't know the vectors. The Hilbert space is built up on metric space and defines new operation as inner product, vectors and using norms. The metric space itself doesn't know norms and vectors, since as is introduced on Hilbert space. Below is a derivation how to utilize metric space in Hilbert space.
Reference: Muscat, J. Functional Analysis An Introduction to Metric Spaces, Hilbert Spaces, and Banach Algebras, Springer, 2014.


What limit does the following inequality involving the sum of divisors and Euler totient functions reach? Consider the following inequality:
$$\frac{6}{{\pi}^2} < \frac{\sigma(n)\phi(n)}{n^2} \lt 1.$$
For the even perfect numbers, this inequality seems to converge to 1. Here's the following ($5$ decimal) values for the first $5$ perfect numbers:
$$0.66666,0.85714,0.96774,0.99212,0.99987$$
Question: Does the convergence to $1$, if it can be proven, show that there are an infinite number of even perfect numbers?

Consider the even perfect numbers. (We don't know if odd ones exist)
By Euclid-Euler Theorem, $n = 2^{p-1} (2^p - 1)$ for some prime $p$ and $2^p-1$.
$$\phi(n)=\phi(2^{p-1})\phi(2^p-1) = 2^{p-2}(2^p-2)$$
$$\frac {\sigma (n) \phi(n)}{n^2} = \frac {(2n)(2^{p-2}(2^p-2))}{n (2^{p-1} (2^p - 1))} = 1-\frac 1{2^p-1}$$
hence this value does converge to $1$ (for the even perfect numbers).
However, as the existence of an infinite number of perfect numbers (or Mersenne primes) is not proven yet, I would presume this argument leads nowhere.

(Note:  This is too long for a comment and complements player3236's answer.)
Consider a hypothetical odd perfect number $N$.
From the following source:
Advanced Problem H-661, On Odd Perfect Numbers, Proposed by J. L´opez Gonz´alez, Madrid, Spain and F. Luca, Mexico (Vol. 45, No. 4, November 2007), Fibonacci Quarterly, we have the bounds
$$\dfrac{240}{217\zeta(3)} < \dfrac{\sigma(N)\varphi(N)}{N^2} < 1,$$
where we have the rational approximation
$$\dfrac{240}{217\zeta(3)} \approx 0.92008188672520641697672952496390495972393334311564395865.$$
Note that we also have
$$\dfrac{6}{{\pi}^2} \approx 0.607927101854026628663276779258365833426152648.$$


A question about convergence test in $2^n$ form For a decreasing and non-negative sequence {$a_n$}, we have:

if $\displaystyle \sum_{k=1}^{\infty}2^k \cdot a_{2^k}$ converges, then $\displaystyle \sum_{k=1}^{\infty}a_{k}$ also converges.

Is the converse of this statement also write? That is:

If $\displaystyle \sum_{k=1}^{\infty}a_{k}$ converges, does $\displaystyle \sum_{k=1}^{\infty}2^k \cdot a_{2^k}$ also converge? And why?

Thank you in advance.


Inequality involving positive real numbers Let $x_1,x_2,\cdots x_n,x_{n+1}$ be any real numbers greater than or equal to $1$.

Then for $n\ge 2,$ I was trying to verify  the validity of the inequality
$$\frac{n-1}{n}\sum_{k=1}^n\frac{1}{1+x_k}+\frac{1}{1+x_{n+1}}+\frac{1}{1+x_1.x_2.\cdots.x_n}\ge \frac{n}{n+1}\sum_{k=1}^{n+1}\frac{1}{1+x_k}+\frac{1}{1+x_1x_2\cdots x_nx_{n+1}}.$$ May I kindly seek your suggestions?

Same result can be tried when $x_1,x_2,\cdots x_n,x_{n+1}$ are non-negative real numbers less than or equal to $1$.

An attempt towards the solution:
We need to prove
$$(n^2-1)\sum_{k=1}^n\frac{1}{1+x_k}+\frac{n(n+1)}{1+x_{n+1}}+\frac{n(n+1)}{1+x_1.x_2.\cdots.x_n}\ge n^2\sum_{k=1}^{n+1}\frac{1}{1+x_k}+\frac{n(n+1)}{1+x_1x_2\cdots x_nx_{n+1}},$$ which is équivalent to $$\frac{n}{1+x_{n+1}}+\frac{n(n+1)}{1+x_1.x_2.\cdots.x_n}-\frac{n(n+1)}{1+x_1x_2\cdots x_nx_{n+1}}-\sum_{k=1}^n\frac{1}{1+x_k}\geq 0.$$
When $x_{n+1}=1$  the result is true. But  for very large values of $x_n,x_{n+1} (x_{n}\rightarrow\infty, x_{n+1}\rightarrow\infty)$ the above inequality is not true.
Hope my argument is correct!

The inequality is equal to

$$\frac{n-1}{n}\Big(\sum_{k=1}^n\frac{1}{1+x_k}\Big)+\frac{1}{1+x_{n+1}}+\frac{1}{1+x_1.x_2.\cdots.x_n} \ge \frac{n}{n+1}\Big(\sum_{k=1}^{n}\frac{1}{1+x_k}\Big)+\frac{n}{n+1}\Big(\frac{1}{1+x_{n+1}}\Big)+\frac{1}{1+x_1x_2\cdots x_nx_{n+1}}$$

We know that $\frac{n-1}{n}>\frac{n}{n+1}$ hence

$$\frac{n-1}{n}\Big(\sum_{k=1}^n\frac{1}{1+x_k}\Big)\ge\frac{n}{n+1}\Big(\sum_{k=1}^{n}\frac{1}{1+x_k}\Big)$$

Since $1>\frac{n}{n+1}$
$$\frac{1}{1+x_{n+1}}\ge\frac{n}{n+1}\Big(\frac{1}{1+x_{n+1}}\Big)$$

Since $a\le b \implies \frac{1}{a} \ge \frac{1}{b}$ hence

$$\frac{1}{1+x_{n+1}}\ge\frac{1}{1+x_1.x_2.\cdots.x_n.x_{n+1}}$$

Adding the three equations above gives us the inequality we seek. $\ _\square$


Are polynomials with the same roots identical? I know that polynomials can be refactored in terms of their roots. However, this must imply that two different polynomials have different roots (this is just what I think). So my question is: Are polynomials with the same roots identical? - if so, why?

A follow-up question that is also about the uniqueness of roots and polynomials can be found here: Is the set of roots unique for each $g(x)$ in $a_n x^n + g(x)$?

No, they are not.

For instance, $2x^2-2$ and $x^2-1$ have the same roots, yet they are not identical.

And, depending on what you mean by "the same roots", we have that $x^2-2x+1$ and $x-1$ have the same roots, yet they are not identical.

Again, depending on what you mean by "the same roots", $x^3+x$ and $x^3+2x$ both only have one real root, yet they are not the same.

However, if two monic polynomials have the same roots, with the same multiplicities, over some algebraicaly closed field (like the complex numbers $\Bbb C$) then yes, they are identical.

Of course NOT. A simple multiplication by a constant works. More interestingly define an equivalence relation where p1~p2 iff they share exactly the same roots!


Side-limit of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ The problem wants the value of left hand limit of the derivative of $f(x)=\frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$ when $x\to3^-$.I graphed the function and the answer is $\frac{3}{2}$ but what's wrong with my method:
For any value $x\to3^-$ we have $[x]=2$ , $|x-1|=x-1$ so if $h\to0^-$, then $f(3+h)=\frac{2(3+h)^2}{2+h}=\frac{2h^2+12h+18}{2+h}$. So the value of the derivative is equal to:
$$\lim_{h\to0^-}\dfrac{f(3+h)-f(3)}{h}=\lim_{h\to0^-}\frac{\frac{2h^2+12h+18}{2+h}-\frac{27}{2}}{h}$$ which is infinity and clearly wrong with or without the graph. I'm not new into this but I can't figure out what's wrong here despite I thought about the method for one hour straight.

I would disagree with you that the answer is 3/2. The left hand limit is +inf, and the right hand limit is -inf, as you can see if you graph the function of the derivative.

$$f(x) = \frac{x^2\cdot\lfloor x\rfloor}{|x-1|}$$
Assume $0 \le \delta < 1$. Then
\begin{align}
   f(3)&= 13.5\\
   f(3-\delta) &= 9 \\
\hline
   f'(3^-)
   &= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &= \lim_{\delta \to 0^+}\dfrac{f(3) - f(3-\delta)}{3 - (3-\delta)} \\
   &= \lim_{\delta \to 0^+}\dfrac{4.5}{\delta} \\
   &= \infty
\end{align}
You did it correctly. Why? Because the floor function is discontinuous from below.
$f(3^-) = 9$ and $f(3) = f(3^+) = 13.5$


Branch of multiple-valued function (complex analysis) Find a branch of $(z^3 - 1)^{1/3}$ which is analytic in $|z| > 1$

So we essentially want to study $\frac 13\text{Log} (1 - \frac 1{z^3} )$, the principal branch of the logarithm where $-\pi < \text{Arg} (z) < \pi $. Problems arise when $\text{Re}(z^3 - 1) \leq 0 $ and $\text{Im} (z^3 - 1) = 0$ i.e. when $z^3 - 1$ is a non-positive, real number.

Setting these equations up, how can I produce the result $|z| > 1$?

Let
$$
\omega_1=1, \quad \omega_2=\frac{-1+i\sqrt{3}}{2}\quad\text{and}\quad
\omega_3=\frac{-1-i\sqrt{3}}{2}.
$$

A domain for $\,\sqrt[3]{1-z^3}\,$ could be
$$
\Omega=\mathbb C\smallsetminus \big([0,\omega_1]\cup[0,\omega_2]\cup [0,\omega_3]\big).
$$
By $[a,b]$, for $a,b\in \mathbb C$ we denote the segment connecting $a$ and $b$ in the complex plane.

Clearly $\{z:\lvert z\rvert>1\}\subset\Omega$.

In order to prove it you need the following lemma:

If $a,b$ belong to the same connected component of $\mathbb C\smallsetminus\Omega$, then
$$
f(z)=\log\left(\frac{z-a}{z-b}\right),
$$
is definable as holomorphic in $\Omega$.

Using this lemma, we can easily see that $\omega_1,\omega_2,\omega_3$ belong to the one and only component of $\mathbb C\smallsetminus\Omega$ and
$$
\sqrt[3]{1-z^3}=(z-\omega_3)\exp\left(
\frac{1}{3}\log\left(\frac{z-\omega_1}{z-\omega_3}\right)+
\frac{1}{3}\log\left(\frac{z-\omega_2}{z-\omega_3}\right)
\right).
$$


Product topology on $X \times Y$ is the coarsest topology on $X \times Y$ that makes the projection maps continuous Show:Product topology on $X \times Y$ is the coarsest topology on $X \times Y$ that makes the projection maps $\pi_x,\pi_y$ on $X \times Y$ continuous.
Attempt: Let $\pi_x:X \times Y \rightarrow X$ and $\pi_y:X \times Y \rightarrow Y$ be the projection maps on $X \times Y$. Let $\tau$ be the product topology on $X \times Y$, and let $\tau'$ be any topology on $X \times Y$ that makes the projection maps continuous. Need to show $\tau \subset \tau'$.
Let $\pi_x,\pi_y$ be continuous. Then by definition of continuity, for any sets $U,V$ open in $X,Y$ respectively,$\pi_x^{-1}(U)=U \times Y,\pi_y^{-1}(V)=X \times V $  are in open in the topology on $X \times Y$. But since $\pi_x^{-1}(U),\pi_y^{-1}(V)$ are necessarily in $\tau$ and in order for the projection functions to be continuous for any topology $\tau'$ on $X \times Y$, $\pi_x^{-1}(U),\pi_y^{-1}(V) \in \tau'$, $\tau \subset \tau'$ and $\tau$ is the coarsest topology on $X\times Y$ making the projection functions continuous.
Comment: Is this all I needed to show for this proof? Is this completely correct, because it seems the proof is mostly trivial, so I am unsure.

The product topology is generated by the subbase $$\mathcal{S}:= \{\pi_X^{-1}[O]\mid O \in \mathcal{T}_X\} \cup \{\pi_Y^{-1}[O]\mid O \in \mathcal{T}_Y\}$$
by definition and if $\tau'$ makes $\pi_X,\pi_Y$ continuous, by definition of contituity it must contain $\mathcal{S}$ and thus $\tau$ (which is the minimal topology containing $\mathcal{S}$..). Ça c'est tout.

Yes, it is that easy. What do you think of the following way of writing the proof?

Let $\tau$ be the product topology on $X \times Y$, and $\tau'$ be any topology on $X \times Y$ such that $\pi_X : (X \times Y,\tau') \to X$ and $\pi_Y : (X \times Y,\tau') \to Y$ are continuous. Let $O_X$ be an open subset of $X$ and $O_Y$ be an open subset of $Y$. Then, $\pi_X^{-1}(O_X) \in \tau'$ and $\pi_Y^{-1}(O_Y) \in \tau'$ by assumption, and so $O_X \times O_Y = \pi_X^{-1}(O_X) \cap \pi_Y^{-1}(O_Y) \in \tau'$. Thus, if $U \in \tau$, $U$ is a union of sets of the form $O_X \times O_Y$, and therefore $U \in \tau'$ by the above argument.


A predicate within a predicate? I'm new here. I wish to ask a question regarding predicate logic:

I was given three predicates:

parent(p,q): p is the parent of q.

female(p): p is a female.

p = q: p and q are the same person.

Now, I was tasked with translating this sentence: Alice has a daughter.

My answer was: There exists a q such that parent(Alice,female(q)).

The answer given is: There exists a q such that female(q) AND parent(Alice,q).

Is it correct to have a predicate (in this case, female) within another predicate (in this case, parent)?

Much appreciated.

Intuitively, a predicate in a predicate doesn't make sense; predicates only take terms as arguments. Using the normal convention of abbreviating predicates and terms by letters ($P(p,q)$ for parent, $F(q)$ for female, $a$ for Alice), your example is
$$\exists q\ P(a,F(q))$$
which is interpreted as "Alice is the parent of true" – absurd, since children aren't truth values. In other words, your attempt does not produce a well-formed formula.

The given answer translates as
$$\exists q\ F(q)\land P(a,q)$$
which is well-formed.

No, it is not in this particular case. You conclude that by substitution.

female(q) can have two values : true or false.

For female q, female(q)=true you have : parent(Alice,true).
For non-female q, female(q)=false you have : parent(Alice,false).

Since false and true are not humans (they are truth values) the proposition parent(Alice,female(q)) is false for all values of q.

So there does not exist a q such that parent(Alice,female(q)).

Your predicate within a predicate is a well-formed formula. True and false can be used as terms of a predicate. But the formula does not express the required concept.


Basic Geometric Series this is very easy I think but I don't have much experience with geometric series. I need to find/simplify the sum

$x^k+x^{k+1}+\dots+x^{k+m}$

where $x\in\mathbb{R}, k\in\mathbb{Z},m\in\mathbb{N}$

So I'm thinking I just write this as

$x^k+x\times x^k+x^2\times x^k+\dots+x^m\times x^k$

Now isn't this a geometric series with $a=x^k$ and $r=x$? So it should have the sum

$$s=\frac{x^k(1-x^{k+m})}{1-x}=\frac{x^k-x^{2k+m}}{1-x}$$

Is this the correct answer?

$$S=x^k+x^{k+1}+\dots+x^{k+m}$$
No need to know by heart some formula. Just multiply by $(1-x)$ :
$$S(1-x)=(x^k+x^{k+1}+\dots+x^{k+m})(1-x)\\=x^k+x^{k+1}+x^{k+2}+\dots+x^{k+m}-x^{k+1}-x^{k+2}-\dots-x^{k+m}-x^{k+m+1}$$
Simplify (this is a telescoping series) :
$$S(1-x)=x^k-x^{k+m+1}$$
$$S=\frac{x^k-x^{k+m+1}}{1-x}$$

The geometric serie  (( 1+x+x^2........+x^{m+1}) has nothing to do with k it doesn't depend on k. It depends on m rather.

The correct term should be $x^{m+1}$ and not $x^{m+k}$, so the geometric serie has the sum:

$$S_m= \frac {X^{m+1} -1}{X-1}$$

And then you have to multiply by $x^k$ to get the whole sum.


If $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$; Find out the no of intersecting point Question from an entrance paper
If $f(x)=x^{2}$ and $g(x)=x \sin x+\cos x$ then
A. $f$ and $g$ agree at no point
B. $f$ and $g$ agree at exactly one point
C. $f$ and $g$ agree at exactly two point
D. $f$ and $g$ agree at more then two point

My approach
$f(x)=x^{2}$
$g(x)=x \sin x+\cos x$
$\Rightarrow f(x)=g(x)$
$\Rightarrow x^{2}=x \sin x+\cos x$
$\Rightarrow x^{2}-x \sin x-\cos x=0$
$\Rightarrow F(x)=0$
$F(0)=-1$ & $F(\pi)=\pi^{2}+1 >0 $
$F^{\prime}(x)=2 x-x \cos x$
$F^{\prime}(x)=x(2-\cos x)$
Here $2-\cos x$ is greater than $0$ and $x(2-\cos x)$ will be also greater than 0 for $x>0$
$\Rightarrow F(x)$ is strictly increasing function.
So I can conclude there will be exactly two point on which they will meet.

$\boxed{\text{*My question is how to do this without calculus*}}$
Edit $1$:- I had thought like this. This a  eq of power $2$,so It must have two roots. These two roots may be real or complex conjugate.Now how to distinguish in general whether it would have real or imaginary roots for all even powered polynomials
ps Dont give solutions which is similar to mine.

A solution:
(we want to show that the equation $x^2-x\sin{x} -\cos{x}=0$ has exactly two roots)
Let $g(x)=x^2-x\sin{x} -\cos{x}$
$g$ is continuous (as the sum of the polynomial $x^2$, $-\cos{x}$ and $x\sin{x}$,which is continuous as multiplication of continuous functions).
$g(-\pi)=\pi^2+1>0,g(0)=-1<0, g(\pi)=\pi^2+1>0$
Because g, is continuous applying Bolzano's theorem: g(x)=0 has at least one solution for $x\in(-\pi,0)$ and one for $x\in(0,\pi)$ Therefore g(x)=0 has at least two solutions.
Let's assume g(x)=0 has more than two solutions, then g(x)=0 has at least three solutions $x_1<x_2<x_3$
$g$ is also differentiable, because of Rolle's theorem  $g'(x)=0$ has at least two different solutions. $g'(x)=0\iff x(2+\cos{x})=0\iff x=0$ so it holds $g'(x)=0$ has only one solution, which leads to contradiction. So g(x)=0 has exactly two solutions $\square$

I don't know if this qualifies as "without calculus." You know that $f(x)$ is monotonously decreasing in $\left. \right] \infty,0 \left] \right.$ and monotonously increasing in $\left[ \right. 0,\infty \left[ \right.$ (you know because you can take the derivative, although in this simple case you don't need to; this is why I'm saying I don't know if this qualifies). You also know that $f(x)$ grows faster than $g(x).$ Finally, you also know that $f(0)=0$ and $g(0)=1$. Putting all this together yields two solutions.


What's a good place to learn Lie groups? Ok so I read the following article the other day: http://www.aimath.org/E8/ and I wanted to learn more about lie groups. Using my exceptional deduction skills I thought "oh it must have something to do with groups" So I picked up a copy of Dummit and Foote's book on abstract algebra and skimmed through it. It didn't say anything about Lie groups however. $E_8$ is coming to be rather famous so maybe other people are interested in this question too. Lets suppose I wanted to learn about lie groups. What books should I read to be ready to learn about Lie groups and what is a good book that talks about Lie groups. I'm guessing its a combination of group theory (representation theory in specific) and also differential geometry. Is this correct? Thank you very much for your time.

I think a good place to start with Lie groups (if you don't know Differential Geometry like me) is Brian Hall's Book Lie Groups, Lie algebras and Representations. The strength of such a book for me would be that it talks about matrix Lie groups, e.g. $SO(n),U(n),GL_n, Sp_n,SL_n$ and not general Lie groups in terms of abstract manifolds. Furthermore, the Lie algebra is introduced not as an abstract linear space with a bracket but as the set of all matrices $X$ such that $e^{tX}$ lands in the matrix Lie group for all $t$.

I am using this book now for a course and I find it extremely readable. For one, proofs are presented in almost complete detail and it is easy to follow. By this I mean one does not need a lot of prerequisites to understand the material. You should of course have an understanding of linear algebra, as well as know topological concepts like connectedness, compactness and path-connectedness.

In conclusion, I think the main strength of Hall's Book is that it teaches you ideas through lots and lots of examples. For example, an entire chapter (IIRC chapter 5) is devoted entirely to the representation theory of the Lie algebra $\mathfrak{sl}_3(\Bbb{C})$. I learned a lot from that example there!

An Introduction to the Lie Theory of One-Parameter Groups
https://www.forgottenbooks.com/.../AnIntroductiontotheLieTheoryofOneParameterGr...
Author: Abraham Cohen; Category: Calculus; Length: 256 Pages; Year: 1911.


Computing an indefinite integral: $\int \frac{2n!\sin x + x^n }{e^x + \sin x + \cos x + P_n (x)}\, dx $ Let $\displaystyle
P_n (x) = 1 + \frac{x}{1!} + \frac{x^2 }{2!} + \cdots + \frac{x^n }{n!}
\
$ and $$
I(x) = \int \frac{2n!\sin x + x^n }{e^x  + \sin x + \cos x + P_n (x)}\, dx
$$ (where $\
n \to \infty
\
$).

This problem is REALLY frustrating to me at the moment, it's 6 AM here and I've been trying to sort it out since 4:30 AM. First of all, I don't get the use of the $P_n(x)$ notation,  isn't that just $e^x$ ?
Anyhow...None of my approaches yielded any useful results, so I'm reaching out to you. Can someone suggest anything, at all ?

It would be much appreciated, thanks a lot!

EDIT : I managed to solve part of it, I'm now stuck with $
I(x) = n!(x - \log [2e^x  + \sin (x) + \cos (x)] + \int {\frac{{x^n }}{{e^x  + \sin x + \cos x + P_n (x)}}} dx
$.

Can't really figure out if this is much better, but that's all I could get up until this point.

This is a tricky integral!

Let's prove that

  $$
\begin{align}
\int \frac{2n!\sin x + x^n }{e^x  + \sin x + \cos x + P_n (x)} {\rm d} x
& = n! \:x-n!\:\log |e^x+\cos x+\sin x+P_n(x)| +C,
\end{align}
$$

where $C$ is any constant (depending on $n$).

Observe that
$$
P_n (x) = 1 + \frac{x}{1!} + \frac{x^2 }{2!} + \cdots + \frac{x^n }{n!}
$$
is such that
$$
P_n '(x) = 1 + \frac{x}{1!} + \frac{x^2 }{2!} + \cdots + \frac{x^{n-1} }{(n-1)!}= P_n (x) -\frac{x^n}{n!}.
$$
Setting
$$
f(x):=e^x+\cos x+\sin x+P_n(x)
$$ we then have
$$
f'(x)=e^x-\sin x+\cos x+P_n(x)-\frac{x^n}{n!}
$$
and
$$
f(x)-f'(x)=2\sin x+\frac{x^n}{n!}
$$
Hence your integral may be rewritten as
$$
\begin{align}
\int \frac{2n!\sin x + x^n }{e^x  + \sin x + \cos x + P_n (x)} {\rm d} x
&= n! \int \frac{f(x)-f'(x) }{f(x)}{\rm d} x \\\\
&= n! \int {\rm d} x-n! \int \frac{f'(x) }{f(x)}{\rm d} x \\\\
& = n! \:x-n!\:\log |f(x)| +C \\\\
& = n! \:x-n!\:\log |e^x+\cos x+\sin x+P_n(x)| +C,
\end{align}
$$ where $C$ is a constant (depending on $n$).

Is the $n\rightarrow\infty$ apply to the $n$ in the integral numerator as well as to $P_{n}(x)$?  If so, then my guess is that either this is an incorrectly posed problem or the answer is totally divergent to infinity due to the $n!$ in the numerator.  I have been working to find ways to get rid of this but the first integral shows that the first part diverges as $n\rightarrow\infty$.


Please prove the following: Given $ƒ(x) = e^x$, verify that $\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$. Given $ƒ(x) = e^x$, verify that
$$\lim_{h\to 0}\frac{e^{x+h} – e^x}{h} = e^x$$  and explain how this illustrates that $f'(x) = \ln e \cdot f(x) = f(x)$.

A proper answer to this question is fully dependent on the proper definition of symbol $e^{x}$. And further a proper definition of $e^{x}$ can not be given by just defining $e$.

One definition of $e^{x}$ is given by $$e^{x} = \lim_{n \to \infty}\left(1 + \frac{x}{n}\right)^{n}\tag{1}$$ Although it is normally self-evident because of the symbol $n$, I want to emphasize that the $n$ in the above definition is a positive integer. Based on this definition it can be proven that $$e^{x + y} = e^{x}\cdot e^{y}\tag{2}$$ and $$\lim_{x \to 0}\frac{e^{x} - 1}{x} = 1\tag{3}$$ Both the equations $(2)$ and $(3)$ are proved in this answer based on definition $(1)$. Using these equations $(2), (3)$ it is easy to prove that $$\frac{d}{dx}\,e^{x} = e^{x}\tag{4}$$ Second part of the question requires us to define the symbol $\ln$ (or $\log$ properly). One such definition is that if $e^{y} = x$ then $y = \ln x$ and then $\ln e = 1$ and this solves the second part of your question.

Using rules of exponentials,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = \lim_{h \rightarrow 0} \frac{e^xe^h-e^x}{h}$$
$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x \lim_{h \rightarrow 0} \frac{e^h-1}{h}$$

This has a limit of $1$ (although I'm not sure how to show this without using L'Hospital's rule. Therefore,

$$\lim_{h \rightarrow 0} \frac{e^{x+h}-e^x}{h} = e^x$$


Why does the name "epimorphism" refer to a surjective homorphism? The wikipedia page talks about epimorphisms with category theory in mind, but I have no experience with this and ask this question from a group theory point of view (answers from any point of view are very much welcome though).

In group theory, an epimorphism is a surjective homomorphism. Where does this name come from? Why have we chosen it?

As a comparison, a monomorphism is an injective homomorphism. Some authors like to call injective functions "one-to-one", and so one can see the rationale behind the "mono" part of the name. But for an epimorphism, I'm not really sure where the terminology has come from (despite reading the wikipedia section on it).

The prefix "epi-" in Greek has several meanings, but a common one is "upon, over". This is similar to the meaning of the prefix "sur-" in French, which was the origin of the term "surjective", introduced by Bourbaki. As such, both give the meaning that the function/morphism "covers" all of its range.

"Epic" has connotations of "long" or "all the way", as well as plenty of others consistent with being onto.

Think of an epic morphism.


Find the integral of $\tan^4(x) \sec(x)$ I want to know if there is a shorter way to find integral of $\tan^4(x) \sec(x)$
without using reduction formula of $\sec(x)$
because it's really takes a long time.

Thanks all

Write it as
$$
\int\frac{\sin^4x}{\cos^5x}\,dx=
\int\frac{\sin^4{x}}{\cos^6x}\cos x\,dx=
\int\frac{\sin^4{x}}{(1-\sin^2x)^3}\cos x\,dx
$$
Can you see an obvious substitution?

What about $(\tan x)^3 \cdot \dfrac{1}{(\cos x)^2} \cdot \sin x$ and make integration by parts?


3D coordinates of circle center given three point on the circle. Given the three coordinates $(x_1, y_1, z_1)$, $(x_2, y_2, z_2)$, $(x_3, y_3, z_3)$ defining a circle in 3D space, how to find the coordinates of the center of the circle $(x_0, y_0, z_0)$?

There are plenty of online articles for the 2D case. A simple google search will show that  this link provides a good explanation about how this is done in 2D. It also shows how to construct the circle center geometrically. So, what you need to do is

1) Find a plane from the 3 points and create a 2D coordinate system on that plane.
2) Convert all 3 points to that 2D coordinate system.
3) Find the circle center using the link above.
4) Convert the circle center (in 2D) back to 3D.

Edit 1: I added the steps for creating a local coordinate system (CS) on a plane defined by 3 points

1) Compute unit vector n1 from P1 and P2. Use this as the x-axis for the local CS.
2) Compute unit vector n2 from P1 and P3.
3) Use n1 x n2 (where 'x' means the cross product) as the z-axis of the local CS.
4) Use (n1 x n2) x n1 as the y-axis of the local CS.
5) Now, you have a local coordinate system, I hope that you know how to convert P1, P2 and P3 to this local CS. After the conversion, the new coordinates for these 3 points should all have their z values = 0.0. You can then use their (x, y) values to find the center of the circle.

If you have all 3 points collinear, you cannot create a local CS and you cannot find a circle from 3 collinear points either.

Let's assume the circle is defined by the 3 points a, b and c, and that the center is denoted d. Following the hints in the answers above the coordinates of points a, b and c can be plugged into the equation of a circle. 3 circle equations and 4 unknowns (3 coordinates of the center c plus the radius) means that we have 2 remaining unknowns for one equation. This equation in fact represents the line passing through the center of the circle, having equal distance to points a, b and c. Intersecting this line with the plane spanned by a, b and c gives the point c. The rest is algebra, making use of some smart substitution of variables:

        Cx = bx-ax
        Cy = by-ay
        Cz = bz-az
        Bx = cx-ax
        By = cy-ay
        Bz = cz-az
        B2 = ax**2-cx**2+ay**2-cy**2+az**2-cz**2
        C2 = ax**2-bx**2+ay**2-by**2+az**2-bz**2

        CByz = Cy*Bz-Cz*By
        CBxz = Cx*Bz-Cz*Bx
        CBxy = Cx*By-Cy*Bx
        ZZ1 = -(Bz-Cz*Bx/Cx)/(By-Cy*Bx/Cx)
        Z01 = -(B2-Bx/Cx*C2)/(2*(By-Cy*Bx/Cx))
        ZZ2 = -(ZZ1*Cy+Cz)/Cx
        Z02 = -(2*Z01*Cy+C2)/(2*Cx)

and finally the coordinates of the center:

        dz = -((Z02-ax)*CByz-(Z01-ay)*CBxz-az*CBxy)/(ZZ2*CByz-ZZ1*CBxz+CBxy)
        dx = ZZ2*dz + Z02
        dy = ZZ1*dz + Z01


Meaning of this symbol What is the mathematical meaning of a horizontal A. I mean that it is like this > but crossed which makes it look like a horizontal A.

Usually, a slash on symbols in mathematics indicates a negation of the original meaning of the symbol.

For example, \begin{align}=\quad&\to\quad\ne\quad(\text{equality})\\\mid\quad&\to\quad\nmid\quad(\text{divisibility})\\\in\quad&\to\quad\notin\quad(\text{element})\\>\quad&\to\quad\ngtr\quad(\text{greater than})\\<\quad&\to\quad\not\lt\quad(\text{less than})\\\sim\quad&\to\quad\nsim\quad(\text{distribution})\end{align}

"≥" This means "equal or greater than"
">" This means "greater than" (being equal is not included)
"≯" This means "no greater than" <= What the hey is the difference between this brat and "≤"???