real analysis – If $f(k) = x^k$ for rational $k$ is continuous at single rational value, will it necessary be continuous for all other rational points?

Suppose $f(k) = x^k$ is a function mapping $mathbb{Q}tomathbb{R}$ where $k$ is rational and $x>0$. If we can prove continuity at a single point in $mathbb{Q}$ is it necessarily the case $f$ will be continuous over all $mathbb{Q}$?

Suppose that single point is taken to be $x’$ with $f : mathbb{Q} to mathbb R$. Then,

$$
forall varepsilon > 0, exists delta > 0 quad text{s.t. with} x’ in D, quad left|x-x’right| < delta quad Rightarrow quad |f(x)-f(x’)| < varepsilon
$$

From here, is it possible to extrapolate the structure and ascertain if it will hold over all $mathbb{Q}$?

real analysis – show that $ f $ is continuous in $a$

If $I = (a,b) $ and suppose $f : I → R $ is increasing in $I$, then $f$ is continuous in $a$ if and only if

$f(a)$ = $inf$ ${f(x):x ∈→ (a,b)}$
I’ve solved it this way, okay? If not, please help me correct it.

Let $f$ continue en $x=a →$ lim $x_→a^+$ $f(x)=f(a)$
Let’s suppose $inf{f(x)/ x ∈ )a,b(}) = L ≠ f(a)→L>f(a)$

lim $_{x→a^+}$ $f(x)=f(a)→$ si $varepsilon = frac{f(a)-L}{2}>0, exists delta$ $> 0 / a<x<a + delta →|f(x)-f(a)|<varepsilon →$
$|f(x)-f(a)|< frac{f(a)-L}{2}→f(x)-f(a) <frac{f(a)-L}{2}→ 2f(x)-2f(a)<f(a)-L→$
$f(x)<frac{3f(a)-L}{2}→ f(x)<frac{3f(a)-f(a)}{2}→f(x)<f(a)$ contradiction then $ x>a$

quadratic forms – Is where a trigonometric field which is different enough from real numbers?

I found this topic in a book ‘Metric Affine Geometry’ by Ernst Snapper and Robert J. Troyer.
I call a field $k$ trigonometric iff there is a quadratic form $q$ over $k^2$ such that every two lines through the origin in $k^2$ is isometric with respect to $q$. This condition is sufficient to introduce trigonometric functions over $mathbf{SO}(k^2,q)$ in a geometric fashion. Hence, a name.

Obviously, $mathbb{R}$ is trigonometric. I know, that to be trigonometric the field $k$ must Pythagorean, that is for every finite sequence of values $(alpha_i)^n_{i=1} in k^n$ there is a $gamma in k$ such that
$$
sum^n_{i=1} alpha_i^2 = gamma^2,
$$

namely every sum of squares is a square. Secondly it must be a formally real field, which means that $-1$ is not a sum of squares. Hence, sadly $mathbb{Q},mathbb{C},mathbb{R}(x),mathbb{Q}_p,mathbb{F}_p$ are all not trigonometric. Probably some extension of $mathbb{R}(x)$ which allows square roots of formally positive functions may work. But I still doubt that it can be totally-ordered, and probably there are some clews in differential Galois theory. Maybe $hat{mathbb{Q}} cap mathbb{R}$, where $hat{mathbb{Q}}$ are algebraic numbers will work, or just adjoining enough real algebraic square roots to mathbb{Q} (call it a Pythagorean closure $overline{mathbb{Q}}$). At least it is Pythagorean and formally real. But I don’t think it is interesting enough.

But I’m very a curious about finding an interesting example of trigonometric field different from $mathbb{R}$. Trigonometric field $k$ different from $mathbb{R}$ may mean formally that $k$ is not between $overline{mathbb{Q}}$ and $mathbb{R}$. I would be very grateful if you could suggest one.
If the result are negative, this would mean that class of all trigonometric fields has certain lower and upper bounds.

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calculus and analysis – Why does this integral of a real, analytic, absolutely integrable function give a complex result?

I am using Mathematica to develop some “interesting” problems for students to solve using Fourier series.

The following computation seems as though it should yield a real result:

$B_n = int_0^1 exp(-9 x^2) cos(n pi x), dx~~ nin Integers,~nge 0$

When I code this in Mathematica, I find it returns a complex result, which does not seem plausible. Here is the code for the first term $(n=0)$ (which is simpler than the general case):

B(0) = Integrate(Exp(-9x^2), {x, 0, 1}, Assumptions -> {x (Element) Reals})

This returns $tfrac{1}{6}sqrt{pi} ,textrm{erf}{(3)}$, which is the correct answer.

However, when I compute the integral for the values of $n>0$, I find the following:

B(n_) = 2 Integrate(Exp(-9x^2) Cos(n Pi x), {x, 0, 1}, 
Assumptions -> {n (Element) Integers && n > 0 &&  x (Element) Reals})

Returns:
$frac{1}{6} sqrt{pi } e^{-frac{1}{36} pi ^2 n^2} left(text{erf}left(3-frac{i pi n}{6}right)+text{erf}left(3+frac{i pi
n}{6}right)right)$

This is a bit baffling. I see no reason that we should have wandered into the complex plane to compute this integral. Anyone have some perspective here?

Thanks.

real analysis – What is the space $D^{k,p}$, and why Sobolev’s embedding holds?

I am reading Struwe’s “Variational methods” and he sometimes uses the space $D^{k,p}(Omega)$, defined as the closure of $C^{infty}_0 (Omega)$ with respect to the norm given by:

$$||u||_{D^{k,p}}^p= sum_{|alpha|=k} ||D^{alpha} u||_p^p $$

The problem is that if $Omega=mathbb{R}^n$, for example, you don’t have Poincaré inequality and thus that should be a different space than $W^{k,p}(Omega)$ (and he uses a different name, indeed). The thing is that he uses embedding results known for Sobolev spaces with functions in $D^{k,p}$. If for instance you look at page 40 he says “By Sobolev’s embedding $D^{k,p} hookrightarrow L^q$ with $frac{1}{q}=frac{1}{p} – frac{k}{n}$“.

So, is there a simple reason for this embedding to hold? Also, do these spaces have a particular name? I couldn’t find anything

real analysis – Let ${x_n}$ be a sequence such that $lim_{n to infty} sup(x_n)=infty$. Then there is a sub sequence of ${x_n}$ that converges to $infty$.

Im not sure how to go about this. I think I am supposed to use the definition of limit superior, but I do not know how to incorporate this. I barely have anything useful, all I have is:

We will start by assuming ${x_n}$ is a sequence and $lim_{n to infty} sup(x_n)=infty$ and we will let $x_{pn}$ be any sub sequence of ${x_n}$.

Can anyone help me, please?

analysis – Can I use geometry to prove that the square root of 2 is real?

For high school students, it is a common exercise to prove that $sqrt2$ is irrational. They assume $sqrt2 = frac{p}{q}$ for some coprime $p$ and $q$, and derive a contradiction.

But being not rational doesn’t mean it’s irrational. It must be also proven that $sqrt2$ is real. Indeed, high school students are aware of imaginary numbers.

It is widely said that to prove $sqrt2$ is real, they must learn real analysis.

But can’t we just use geometry? I could argue the following:

By Pythagoras’ Law, $sqrt2 = sqrt{1^2+1^2}$ is the length of a diagonal of the unit square. Since a length is a positive real number, $sqrt2$ is real.

To think about it, it seems hard to build geometry from the standard axioms, namely ZFC. Does that mean the argument above is just a harder way of a proof?

python – Turn mannequin image to real human image

I have a task at hand, where i am getting a mannequin images which i have to convert into real images so that it can be uploaded on e-commerce.

Here is the sample image i want to convert this mannequin to real images which consist of real face and body.

I am Open for Ideas, i am struggling to think how to achieve this.
I am very well versed in Computer Vision and Deep learning.
Can anybody guide me to the way or method i could use to solve this.