taeto

22 hours ago, gib65 said:

And if you allow \(n = \infty\), then...

As studiot points out, you don't. Or rather, you shouldn't. 

The symbol \(\infty\) is used in various roles in math. Most often as part of a standard notation, such as in \(\sum_{i=1}^\infty\) which is shorthand for \(\lim_{n\to \infty} \sum_{i=1}^n,\) which is itself a shorthand for a more complicated expression that defines the value of such a limit (and which happens to no longer contain the \(\infty\) symbol). In this way the use of \(\infty\) in the expression \(\sum_{i=1}^\infty\) is not much different from the use of the Greek letter \(\Sigma\) in the same expression. It doesn't `stand for' anything, in the same way that \(n\) does in the similar looking \(\sum_{i=1}^n,\) where \(n\) represents a natural number value. 

There are several contexts when \(\infty\) is the symbol chosen to represent some fixed object of particular importance. Most famously when extending the reals \(\mathbb{R}\) by declaring that \(\infty\) be some object that is not itself a real, and that the usual ordering \(<\) of \( \mathbb{R}\) get extended so that \(\infty\) is larger than each real number in a new larger ordering.

At least a dozen other contexts have the symbol \(\infty\) representing objects of special properties. It means that next time you board a plane and sit down in your assigned seat next to a person completely unknown to you, and you think that you should initiate conversation by making some observations on the properties of \(\infty,\) you do have to be specific about the context. If the other fellow happens to be specialist in the theory of elliptic curves, they may not realize you are talking about the \(\infty\) which is used to compactify the complex plane, and they might think, judging from your remarks, that you are a bit weird.

Everyday language has the same ambiguities, where the same word may have different meanings. Cf. the quote from Mae West: "Women like a man with a past, but they prefer a man with a present." And other examples exist. I am not aware of any `language crackpots' who get irate about this fact and are willing to denounce the use of language as defective and useless. In contrast, math crackpots have a field day with examples such as \(\infty\) that can take different meanings depending on context. 

The point is that if your context of choice already assumes that \(\infty\) is the name of a special object, then for all we know it might be that \(\infty\) is fixed at the natural number value \(42.\) In that case you seem justified to let \(n = \infty,\) in which case follows \(\sum_{i=1}^\infty a_i = \sum_{i=1}^{42} a_i.\) This is bad, because then you are using the expression \(\sum_{i=1}^\infty\) in two incompatible ways, which makes it `ill-defined', we would not know what it means, and therefore it means nothing. 

25 minutes ago, geordief said:

Is the Cotangent Space not perpendicular to the Tangent space then ?

If it is not then it seems I have my wires crossed 

I thought it was a space defined by/including  a vector that was erected normal to the point p on a manifold (I was thinking of a 2 dimensional surface)

Perhaps you are telling me that it  is simply the dual space of the Tangent space and there is  no visually geometric representation of it (as there seems to be with the Tangent space)

I am not really familiar with the concept of the Tangent space of a line .....

But you are okay with 2-dimensional examples of manifolds embedded in 3-dimensional space, and in 4-dimensional space?

In any case you have to distinguish between abstract manifolds, which do not always have to exist as a part of any larger spaces, and 2-dimensional surfaces that can be considered to be part of 3-dimensional space. The abstract manifolds do not really have or need any space around them to wriggle around in, and neither do their tangent spaces nor their cotangent spaces.

Then instead we can consider the 2-dimensional manifold \(\mathbb{R}^2\) embedded in our familiar 3-space, as a fairly boring, yet for the purpose still illustrative example. 

For a point \(x\) in  \(\mathbb{R}^2\) do you see what is the tangent space \(T_x\) and the cotangent space \(T_x^*\)? What are their respective dimensions? Is either of them naturally contained as a subspace of the same 3-dimensional space in which  the original \(\mathbb{R}^2\) lives?

4 minutes ago, geordief said:

Would the Tangent space at x be   a plane above and along the line   and the the Tangent space any one of any number of  those vertical "revolving doors"  you go through going into e.g. a hospital?

That "revolving door" is centred at x.

If you take the Tangent Plane and "swing  it" 90 degrees  ( about a fulcrum at x) you also get one of the Cotangent planes(set those planes "revolving" and you get  any number of them.)

Does that make sense?

First of all, since \(\mathbb{R}\) is a 1-dimensional real manifold, the tangent space at any point is a 1-dimensional space, and so it is a line, not a plane.

And in addition, in usual understanding there is only one cotangent space at every point of the manifold. If you have a different understanding which allows for several cotangents at a point, then you have to provide that definition of "cotangent" which you find suitable to allow for this.  

Well, if we look at an actual example, then the real line \(\mathbb{R}\) itself is quite a respectable example of a real manifold, with which I hope you will agree. At any point \(x\) the tangent space \(T_x\) is again isomorphic to \(\mathbb{R}\), right? How would you explain the cotangent space at \(x\), and the geometric picture that you will use to describe it? Is there some sense in which it is perpendicular to \(T_x\)?  

When you have a manifold and a particular point on it, there is a particular tangent space to the manifold located at that point. Now usually the cotangent space to the manifold located at the same point is defined to be the dual space of that tangent space. Your question suggests that you have in mind some other definition of the cotangent space, and if so, then which?

2 hours ago, 113 said:

How many terms should I grab to go safe for every case? Why doesn't it suffice to take just the 1st non-zero term? "

What I was asing in my first post, is it possible to plug in an infinitesimal value? Is it possible to calculate limits using infinitesimals?

You are thinking of a function given as a series expansion with terms ordered from lower to higher orders. But a function in general may not easily be represented in this way, and it may not even be possible. In some cases there maybe does not exist any first non-zero term to just grab.

Formally to find \( \lim_{x\to 0} f(x) \) using infinitesimals, you should consider the exact value \( f(\Delta x), \) as you suggest, by plugging in the infinitesimal value \(\Delta x.\) Once that is done, and if \(f(\Delta x) \) is not infinite, you have to compute the standard part function st\( (f(\Delta x) )\) to get the real value of the limit. How to do that in practice is another story.

If you do happen to know a series expansion of your function, then you do not really need any computation, since the answer depends on whether there are non-zero terms with negative powers, in which case the limit does not exist, and otherwise you read off the constant term. 

1 hour ago, Edgard Neuman said:

For instance, with : 

 x[n+1]=r * x[n] * (1-x[n])

With r = sqrt(2)*2.6  for instance

X[n] is always in [0 ; 1]
How do you find y in [0 ; 1 ] that is not in  { x[n] } ?

Okay then, I will attempt to give the standard answer. I assume \(x(0) \neq 0\) and \(x(0)\neq 1\) as otherwise the answer is trivial, say with taking  \(y= 1/2.\)

The key is to realize that every real number in \([0; 1]\) has a representation, a "name" if you will, as a binary string preceded by a zero and a "decimal" point, such as the representation of the fraction \(3/4\) as the string \(0.11,\) meaning \(1/2\) for the first digit after the decimal point added to \(1/4\) for the second digit. The fraction \(1/3\) has a representation which looks like \(0.10101010\ldots\) etc., hence the representation is not of finite length for every number.

You say that your \(x(n)\) is always in \([0,1],\) for every \(n.\)  In which case it has a representation as described. 

Now we construct a number \(y\) in \([0,1]\) that has a representation which is different from the representation of any one of the numbers \(x(n).\)

For each \(n = 1,2,3,\ldots \) let the \(n\)'th binary digit of \(y\) be the one that is different from the \(n\)'th binary digit of your \(x(n).\)

Then the \(y\) just constructed cannot be equal to any one of the numbers \(x(n),\) since in the \(n\)'th digit it is different.  

1 hour ago, swansont said:

There are biologists out there who believe in creationism. Probably geologists, too, who think there was a worldwide flood.

I think perhaps you can answer the exam questions by knowing the answer "they" want without actually believing it, or perhaps not understanding the ramifications, and still get a passing grade — and with it, a diploma.

There are mathematicians out there too of similar caliber. Which ought be more astounding, given the fact that in the experimental sciences like biology and physics, maybe chemistry to a lesser degree, the student is usually told second-hand about experiences. Whereas in mathematics, all the student has to do is make a look-up in a book to see that the crackpot is not telling the true story.

The question is how as a student you will get equipped to answer the questions convincingly if you both know the actual answers, and you also know the different answers that they will want. 

The Fourier transform of an even real function is real. More generally, it is real for the function \(f\) if \(f(-x) = f(x)\) holds for almost every \(x \in \mathbb{R}.\)

The Fourier transform of a real function is otherwise not itself a real-valued function.

16 hours ago, Edgard Neuman said:

how can we build these numbers ? 

What do you mean by "build these numbers"?

For any sequence we can easily construct an element of the interval \([0,1]\) that is not in the given sequence. This is standard, and I can guess that you most likely know the answer already.  

Are you asking whether it is somehow possible to construct all the elements of \([0,1]\) that are not in the sequence? 

5 minutes ago, dimreepr said:

maybe she's just cross.

A cute Persian cross kitten.

50 minutes ago, Curious layman said:

What you talking about, she's beautiful 😘

feast your eyes on princess Zahra Khanom Tadj es-Saltaneh (1883 – 1936) who was considered the ultimate symbol of beauty in Persia during the early 1900s. So much in fact, a total of 13 young men killed themselves because she rejected their love.

OK, I'll reconsider. If she was 18 at the time of the photo, that might yet give the brows enough of a chance to get their act together. 

On 9/4/2018 at 8:00 PM, koti said:

https://www.sadanduseless.com/beauty-standards/

Not enough unibrow for my taste, so I have to pass .

12 hours ago, mathematic said:

Rather than L'Hopital's, one can take the first few terms of sin(x)=x−x3/3!+x5/5!+... to get -1/3!, as x→0 .

It depends on the view you take of the Taylor series expansion of \(\sin (x)\) around \(x =0.\)

Maybe you would think of \(\sin(x)\) as the imaginary part of \(\exp(ix)\) with \(\exp\) defined by its well-known series expansion around \(0\), in which case your explanation is fine. But that might appear quite mysterious to someone who only knows \(\exp\) as the inverse of the \(\log\) function. Which is how \(\exp\) got reasonably explained back when I went to school.

If not, then to get the third coefficient in the Taylor expansion of \(\sin,\) isn't that just basically just the value of the limit expression in the OP? And then your answer just seems the same as saying that to calculate the limit, you can use the traditional method. Which is not what the OP wanted.

36 minutes ago, Mordred said:

Little trivia side note the reason to square the energy term is to denote positive norm. Ie no negative energy allowed. 

Is there a better way to phrase what this is supposed to mean? We notice that the identity only specifies the value \(E^2.\) Surely this quite definitely allows for two different final solutions for \(E\) itself in the nonzero case; one positive and one negative. 

This very limit got worked out on:

math.stackexchange.com/questions/2487889/solve-this-limit-using-equivalent-infinitesimals .

However, the method has the tiny disadvantage that in order to compute the coefficients of the Taylor Series of the expression in the numerator of the fraction, which you need to know, you have to first produce the value given by the limit anyway , I suppose by ordinary means without using infinitesimals?! EDIT: well, after all, the coefficients are values of the iterated derivatives of \(\sin \) at \(0,\) which can be done symbolically using infinitesimals, so that part is fine.

Btw, it is either L'Hôpital or L'Hospital to you .

7 minutes ago, uncool said:

Let me put it quite simply: no, the process of taking the derivative never involves a division by 0. Period. 

He seems to sense that a vertical line does not have a finite slope. Give him credit for that.

10 minutes ago, Conjurer said:

The derivative of a line intersects a point on the curve, because the height or distance between the two points in question is zero. 

You have an education degree right? What is your answer if a student asks "which two points in question"?

1 minute ago, Conjurer said:

The equation y = 1 or y = constant is excluded from this working.  It is a vertical line, so it fails the vertical line test, and it is not a function.  It has to be a function.

It is still a line though. If you draw it in the xy-plane it even looks suspiciously like a horizontal line, not a vertical line. 

7 minutes ago, Conjurer said:

The derivative is the equation of the line that intersect a curve of a function at exactly one point. 

If that were the case, then \(y = 1\) would be a derivative of the function that maps \(x\) to \(\sqrt{x}\) for \(x \geq 0. \) Since the line with equation \(y=1\) intersects the graph of this function in exactly one point \((1,1)\).

3 minutes ago, Conjurer said:

It is a necessary step to finding the instantaneous value of anything in Newtonian Physics.  This is like asking if Newton knew how to do Newtonian Physics.

If it is really necessary, it should be so much easier for you to present a single example of such a calculation, even without having to go all the way back to the 17'th century. Can you do this? Remember that I am asking about dividing by zero. Dividing by an infinitesimal is no problem, each of us who have been following this thread already knows how to do that. 

17 minutes ago, studiot said:

Philosophae Naturalis Principia Mathematica

Fair enough. This is the first time I heard it referred to as just Principia Mathematica. Thanks for pointing it out Studiot, fist bumps and shake hands all around, I hope. My bad.

Then the claim stands, I suppose: Newton did know how to make sense of division by zero? Somewhere in one of the three volumes and on some page?

9 minutes ago, Conjurer said:

It is ridiculous that I would even have to provide this on a science forum, since it is just the basic proof of a derivative.

Why do you think that you would have to provide the definition of a derivative, if that is what you mean by "proof of a derivative"? Nobody asked you about that. Instead you were asked to provide some kind of reference or evidence of your claim that Newton could make sense of division by zero. Maybe it would help if you just admit that you are simply making up nonsensical stuff all the time.

8 hours ago, Conjurer said:

It was proved by Isaac Newton in the Principia Mathematica.

That cannot be entirely accurate. When Principia was written, Newton had long since passed on. Anyway, if that is where you got it from, then surely you can provide the relevant volume and page number? 

2 minutes ago, studiot said:

Talking about the 'standard' part is no more justifiable than choosing only to take the real part of a complex number.

About that point, I think that the "standard part" of a hyperreal is only defined to be the nearest real number if the hyperreal is finite. In the context of derivatives that just means that if the hyperreal differential quotient is infinite, then you conclude that the derivative does not exist.

17 minutes ago, Conjurer said:

It is mathematically correct to divide by an infinitesimal in this situation, because it has been mathematically proven that you can divide by an infinitely small number or zero in only this one specific situation.

I suspect that according to the rules of the forum, since you state here that something has been proven, then you have to provide some kind of reference or other solid evidence. Is this alleged fact something which has been proven by you personally?