## 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 $$mathbb{Q}$$?

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## 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

$$f(a)$$ = $$inf$$ $${f(x):x ∈→ (a,b)}$$

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 $$varepsilon = frac{f(a)-L}{2}>0, exists delta$$ $$> 0 / a
$$|f(x)-f(a)|< frac{f(a)-L}{2}→f(x)-f(a)
$$f(x) contradiction then $$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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## Follow as I build a real makeup Shopify store from scratch

I’m a website designer and developer, I’ve built commercial e-commerce stores before and I’m always asked “how do you create an online store?” thinking they need to hire me or unsure where to start.

So I created a full tutorial video aimed at people who use the Internet, social media, etc. regularly but not technical or coders themselves. I walk you through creating a real online makeup store on Shopify from scratch while giving tips, tricks and advice from my previous experience of building e-commerce websites. Jack.

## 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.

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## 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

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## 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 $${x_n}$$.

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## 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. 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.