taeto

On 5/16/2018 at 6:14 PM, theodorenghiem said:

Thanks for your feedback.

My second attempt to consolidate 3 rules into one whole number = prime + fibonacci1 + fibonacci2 has  been verified with 10,000,000 without failure. Please view my txt file in the link

 

I feel dyslectic. There is a conjecture suggesting that every natural number \( \geq 2 \) can be written as a sum of a prime and two Fibonacci numbers, is that right?

1 hour ago, MathsLearner123 said:

 

I have attempted the problem. Can you please advise if i am doing correctly. I want to prove equation 1 again by induction is it correct? Please advise.

Induction.pdf

It is a good attempt, though only about half right.

You should not write "for n = n+1", since obviously n is always different from n+1, except in very peculiar situations, which do not appear in ordinary arithmetic. Instead you should write something like "Assume that the statement is true for n, and consider the statement for n+1" (which means, the statement about n+1 that you get after replacing n by n+1 in the original statement).

So you will assume that you have some number n for which you already know \( 2^n \geq n^2 \). The only thing left to show now is \( 2^{n+1} \geq (n+1)^2 \) and you will be done.

You should do it by first grabbing the thing \( 2^{n+1} \) that you want to prove something about (namely that it is \( \geq (n+1)^2 \) ). But you should look at the statement that you know that you are allowed to assume, which is   \( 2^n \geq n^2 \). You have to ask yourself, what does the fact \( 2^n \geq n^2 \) tell me about how large the number  \( 2^{n+1} \) will be? And then, if you can figure this out, how does your new guaranteed smallest possible value for \( 2^{n+1} \), which you just got, compare to the number \( (n+1)^2 \)? Is it a larger number? If it is, then you are done. 

The kind of statement here seems tricky because it involves an inequality instead of just a formula with an equality sign in it. Maybe you should practice a bit more with arithmetical theorems that can be written as equations. Something like \( 1 + 2 + \cdots + n = \frac{n(n+1)}{2} \) is another standard exercise.

You are getting somewhere:

1 hour ago, MathsLearner123 said:

For n=2

LHS: 2 * (9 + 16)  = 50 RHS: 7^2 = 49

50 > 49. Difference (50 - 49) = 1;

is essentially the basis of the induction proof. You have shown that the inequality is true when n is equal to 2. This is important. However,

1 hour ago, MathsLearner123 said:

For n=3;

LHS: 4*(27+64) = 364, RHS: 7 ^ 3 = 343

364 > 343. Differnce (364 - 343) = 21.

is not part of the proof by induction. It shows that the inequality is true when \( n \) is equal to 3. The increase in the difference between LHS and RHS is irrelevant. How would you argue that the statement is also true for \( n = 4, \) except by calculating both sides of the inequality for that case as well? You do not know whether the difference between LHS and RHS will suddenly drop enough to make the inequality false for that value of \( n, \) or for some larger value.

Did you read the Wikipedia page, or anything else, on how to prove something by induction?

Do you know how to prove that \( 2^n \geq n^2 \) holds for every \( n > 3 \) by induction? You should be familiar with this baby example if you want to prove anything as complicated as the problem that you posted. I am puzzled if you are asked to prove anything complicated like that without having seen the simplest examples. Where did you get your problem from? Is there anything to suggest that you should actually try to solve it using Taylor series? 

10 hours ago, fiveworlds said:

That's nonsense there is no proof that ZFC is versatile enough to write p vs np in either direction. P vs np details a set of problems which you have to prove work on a deterministic Turing machine in a polynomial amount of time. For instance I can prove that I can replace words in sentences with a deterministic Turing machine in polynomial time using the following python function which replicates the action of a real deterministic Turing machine. 

But it can do more than that you can easily replace cities "Cork, London, New York, Brussels" with their shortest travelling salesman result. There is of course a limitation of the number of cities but it'll work for flights and stuff.

Did you know that time complexity on a Turing machine is measured by how many moves the read/write head has to make. Therefore if you make the read/write head large enough to read any input on your tape you never need to move the read head anything but forward making all algorithms O(n). Of course there is a problem with that approach because a Turing machine allows for an infinite tape size and it is not possible to have an infinitely large read head. So you could prove p=np simply by showing that no problem requires an infinite amount of tape.

 

 

Sorry, but the rules for the Clay prizes state explicitly that whatever proof you have, it has to be a valid proof in ZFC.

Your last remarks seems not quite on point. Depending on what you mean exactly by your condition of requiring an infinite amount of tape. If a problem is in NP, then it follows easily that each YES-instance of it can be solved without using an infinite amount of tape, simply because there has to be some way to find a solution in finite time. For some NO-instances  of NP problems you have to use infinite time for searching, and I doubt that it has any bearing on anything whether you also use infinite tape or not. But even if the problem is in P, you cannot in general do with just a finite amount of tape, because there will be instances that require more space just for stating their input. Either way, the TM that solves all instances has to have an infinite amount of tape available.

8 hours ago, fiveworlds said:

Actually I will prove you cannot use ZFC to solve p vs np. 

A turing machine reads a two number tape {a,b} and gets the power.

a^b does not equal b^a therefore {1,3} and {3,1} are not equal sets and ZFC cannot be used to solve p vs np.

en.wikipedia.org/wiki/Turing_machine#Formal_definition gives the formal definition of a TM as a set in Set Theory.

Btw it is possible to represent an ordered pair (a,b) as a set as well: the most famous definitions are by Wiener, Hausdorff and Kuratowski, see en.wikipedia.org/wiki/Ordered_pair

51 minutes ago, StringJunky said:

It is what it is. This is not  really a computer-based forum. 

Sure. And sorry about the rant. I am newbie and just hoped that in a science forum it would be easier to find those contributions that actually deal with science topics.

It is possible to use Taylor series to get a proof even in the more general case when n is any real number greater than or equal to 2. But it looks like it will be complicated. Do you know the Taylor expansion of the RHS of your inequality, that is, the expansion of 7^x as a function of x?

It looks more like you are supposed to apply induction on n: en.wikipedia.org/wiki/Mathematical_induction

The easiest might be to first show that the result holds when n = 2, and then prove with simple calculus that the difference LHS - RHS is always increasing for larger n (again assuming the variable is real-valued). This would be kind of "induction on real numbers".

It may no longer be useful, but I will try to add some suggestions. Sorry that I cannot make Latex work on this site.

It is clear that the sum of 1/n! for n=3,4,... is equal to e - 5/2, since we are missing just the terms for n=0,1,2 in getting the actual expansion of e as the sum of 1/n! over all natural numbers n. We just need to work out the sum of the terms of the form 1/(n! n) over n=3,4,... in your final RHS expression. Call that sum S.

Let a function f be defined by f(x) = sum{ xⁿ/(n! n) : n=1,2,... }. Then it is clear that S = f(1) - 5/4, since S is of the form of f(1), except for missing the terms with n=1 and n=2. Also f(0)=0 trivially holds.

Each term  xⁿ/(n! n) in f is the integral of t^n-1/n! dt  from t=0 to t=x. By a simple analysis argument it follows that f(x) is equal to value of the integral of (e^t - 1 - t - t²/2)/t from t=0 to t=x, which is the same as the integral of (e^t - 1)/t - 1 - t/2 from t=0 to t=x. 

Your sum S becomes the value of the integral of (e^t - 1)/t -1 - t/2 from t=0 to t=1 and then subtracting 5/4. And your sum total becomes e - 5/2 - S.   

On 5/19/2017 at 1:12 PM, sevensixtwo said:

Here is the paper: http://vixra.org/abs/1703.0073

Also testing if the "lol XD you so fahnny posting ou schizo viXra maymays xD XD XD" people are active in this forum.

The people who are capable of recognizing when a paper about the Riemann zeta function does not even define the Riemann zeta function properly, and otherwise consists purely of word salad anyway, might however be active in the forum.

On 6/4/2018 at 6:57 PM, John Kenneth Swinswood said:

Pi = circumference divided by diameter. Simple geometry. 

e^x is a function equal to its own derivative. In the physical realm e^kt represents exponential growth if k is positive, or exponential decay if k is negative. No connection with geometry or the circle. Therefore no connection with pi.

Now introduce i.

e^ix is a complex number which leads to a circle on the Argand diagram, and thus to pi.

Without i, no connection between pi and e. Only with i can the connection be made. Magic.

 

In the study of physical oscillating systems, with possible dampening, you naturally obtain second order linear differential equations for their movement, as a consequence of Newton's second law F=ma, where F is force, m is mass and a is acceleration (the second derivative of the position as a function of time). The basic example is a ball suspended on a spring while subject to a viscose environment such as a surrounding air or liquid.

The solutions are products of trigonometric functions like sin(at+b) and cos(at+b) at time t, that provide the oscillations, with exponential functions e^-wt , that provide the dampening effect. This is not magical, just consequences of the fact that the trigonometric functions are very closely related to the complex exponential function. The exponential function and the  trigonometric functions are also the basic functions for which their second derivatives are proportional to themselves. Which is why they appear naturally together in descriptions of how physical systems behave. 

The precise identity e^it = cos(t) + i sin(t) does the rest to establish a connection between e and pi.

This is a science forum. I go to check what goes on in Computer Science, and it will say "Computer Help". Like what is that supposed to mean: there is discussion on how to write programs, or how to repair a desktop? Judging from the titles of contributions, that is actually pretty much it. No trace of science here.

I am not a computer scientist, but I do happen to know what Computer Science is, in real life. namely CS is a mathematical theory which deals with measuring the theoretical efficiency of computation. For example, the prominent open questions are problems like NP=P? and whether natural numbers can be factorized in polynomial time. But there are many more less famous.

For NP=P, if you propose a solution to win the $1M Clay prize money, you have to work out your precise solution in the framework of ZFC set theory, or you will be disqualified. The reason being that ZFC contains the objects that have the structures of Universal Turing Machines, which are ultimately the things the statement NP=P says something about. You cannot just go work from intuition, or say that your new motherboard seems fast, so it will probably do it fine.

CS is probably closer to set theory than to anything else in the forum. I would prefer to have it placed in the corresponding section. Then "Computer Help" could be fine with staying right where it is. I see no connection whatsoever with the notion of "Computer Help" and science, but if the Gods who decide things around here think that it has to stay, then so be it. Just please, whatever it is, do not confuse it with actual CS science.   

The correct spelling is "Schwarzschild" though

Thanks for the formatting tip; I have been too much in time constraints to notice the features.

I am not sure about disingenuity. I am happy as long as the domain is precisely defined, in the shape of a set. Doesn't matter whether sets of numbers or not, those are just the basic examples. To consider other sets does not present any further formal complexity.

My conclusion so far is this. For the typical informal explanation of a function you would have something like: let f : A -> B be the function given by f(x) = (something or other), where something can be anything that starting from an element x of A determines a unique element of B. Even when it is obtained from an application of the Axiom of Choice (though in that case the exact wording of the explanation probably is different). Then there is no problem with identifying the well-known properties; injectiveness, surjectiveness, bijectiveness, and for heck's sake, throw in continuity, differentiability, smoothness etc. as well, supposing they apply.

But for the formal explanation of a function I am not so sure. Q: Is the function x²+3 surjective? A: I don't know, what is your arithmetic? Q: natural numbers. A: no it isn't, you don't get 5 into your range. Q: wait, I meant rationals. A: no, you still don't get 5. Q: wait, I actually meant real numbers. A: the domain is all real numbers? Q: yes. A: what is your co-domain, also the real numbers? Q: yes,of course. A: no, then you do not get 2, please choose a different co-domain. Q: no, instead I want to choose a different domain now: all algebraic complex numbers, how about that? A: and the co-domain is still all real numbers? Q: yes, of course, when did I say otherwise, don't be a jerk. A: no, sorry, that does not even give you a function... And so this exchange can go on forever. Much like conversations in this forum, if I may make that observation. I hope you get the point that so long as you do not have a clear concept in mind of the meaning of a term, like "surjective", and it has to be explained relative to a lot of other circumstances that are arbitrary to the parties, then you are not getting any dialogue established.

But apparently I cannot pinpoint the formalization of "function" which avoids this. 

12 hours ago, studiot said:

I thought I had pointed out that pairs may not be adequate.

Nor need a function be analytic or specified by an equation.

All that is required is that every time you apply it to a member of the domain set you get the same answer.
So it can't be a table of experimental results directly. But it could be a specific table such as a logic truth table (Cayley table)
Or it could be a filter for instance one which has all the integers as the domain but selects only odd integers.

If you are thinking about functions of more than variable, then it seems fine to consider, say, a function of three real variables as a function of a single variable but defined on R^3. More generally a function of several variables can be viewed as a function of a single variable with a domain consisting of the suitable product of sets. In this sense the representation using just pairs (x,y) is quite adequate.

Otherwise I agree with what you say.

6 minutes ago, studiot said:

Specification of the Domain

Specification of the Co-domain

Definition of the rule of mapping.

That seems a standard way of saying it alright. But it is not what I mean. 

I would almost prefer the way in which uncool puts it, because now in your suggestion you get into a problem with this "rule" notion. I can concede that if by "rule" you simply mean: the rule is that y = f(x) holds if (x,y) is the point on the graph of f that has x as its first component. Then that might seem fine as a rule, except it appears quite roundabout, besides being cyclic. But be aware that this "rule" thing can never be interpreted as saying that you have a "formula" for computing y once you are given x as input. Once you apply the Axiom of Choice you will be 100% sure to get heaps of functions none of which can be computed by any formula.   

Now I try a definition that makes sense to me.

A function f is a pair f = (F,B), for which there is a set A such that F is a set of pairs (x,y) with x in A and y in B, such that for every x in A there is precisely one pair in F with x as its first component. If (x,y) is this pair then we write y = f(x). The set A is called domain of f, the set B is called co-domain of f, and F is called the graph of f. The range f(A) of f is the set of y in B for which there exists x in A with (x,y) in F.  We say that f is injective if (x1,y) and (x2,y) both in F implies x1 = x2, and that f is surjective if f(A) = B, and that f is bijective, if f is both injective and surjective.

I think that would be alright in set theory, unless I missed something. But it is horribly cumbersome. I can imagine that you agree at least on this point. Have you seen it done better, while roughly capturing the same meaning?

27 minutes ago, studiot said:

There is no such thing in general as the co-domain.

A valid co-domain is any set that includes all the values of a function.

That honour is given to a more restricted set called the range which is the set of values taken on by a function as its argument varies through its domain.
Thus the range is the image of the domain in the co-domain.

Now if every member of the of the co-domain is the image of at least one member of the domain then the function is said to be surjective.
In this case the range and the co-domain are the same or coincident.

As an example the mapping from the set of all men to the set of married women is surjective (modern gender relations aside). But there remain men in the domain who are not married, that is their function value is null.

It's tricky to get all the definitions to match sweetly, which is why so much thought was put into it by many clever people.

How are we doing?

Brilliantly, I think! Thanks for your input.

You have a different understanding from me of the properties of a function, judging from your example. Are you certain that you would argue that the function that maps x to 1/x has the entire set of real numbers as its domain, and it just happens to be the case that there is an element 0 in the domain that has null function value? That seems controversial, and not the usual understanding.

And you suggest that a co-domain is not (necessarily) fixed in the description of a function. In which case it is moot to ask about surjectivity, because every function is surjective if only you choose its co-domain small enough. Also bijectiveness would be equivalent to injectiveness, by the same token. I can agree that this is also what I get from the raw set-theoretic definition of a function stated by uncool. It still does not seem right; it should be possible to ask questions about surjective versus non-surjective. I just do not see how to best do it in the way of a simple explanation, or how to do it exactly formally either. 

1 minute ago, studiot said:

 

That was not your question, although it would have been addressed, but I note that you don't want a discussion on that subject.

Note also that the range is a different thing from the co-domain.

     I don't know what you mean; this was exactly the first part of my question: how do you define "function" in a way so that you can identify both "what it does" and also whether it has a property like being surjective, which was in the question by the original OP. The fact that the range is not the codomain is obviously the important reason why you cannot do that from the graph of the function alone. 

What is it that you think that I do not want a discussion about? Sorry to seem dismissive about your thoughts on singled-valuedness, I just did not think it relevant.

2 minutes ago, uncool said:

In terms of set theory, a function is a set of ordered pairs (x, y) such that for each x, there is a some unique y such that (x, y) is in the set. 

      It is indeed. But then how do you answer the question about surjectiveness? You can assume that the range is known; the set of all y that appear as values. How do you identify the codomain?

28 minutes ago, studiot said:

This is a very good subject to consider as it highlights an important property of functions, that of being single valued.

Consider the function

f(x):x=4–√

Now the point about this very simple example is that 4 has two square roots, -2 and +2.

So if we allowed the function we call the square root function to have two values then we could justifiably say that for f(x) :  x + x = 0, but that f)x): 2x = +4 or -4.

 

We can develop this further if you like and also investigate other properties of functions.

       There should not be any worry about functions being single valued. If you use a square root notation, it is automatically assumed that you mean the principal root, which is nonnegative in the real case, in particular. That is not the point.Other than that, your example is better than mine, since if you take a function to mean a "formula", then assigning the value √4 to x still seems different from assigning the value 2 to x, since the "formulas" for the assigned values are different.

      You use the term "function". What do _you_ mean exactly by this notion? That was my query, and I think also in the spirit of the OP's original question. 

I was drawn to the title of this somewhat dated thread.

Apparently to deduce the properties of a function you have to know not only "what it does", but also its exact domain and codomain; the sets on which the function is defined, and the set in which its values are found, respectively. You might think that the most efficient method of "knowing" everything about a function F is to have a complete table of the values of (x,y) for which y = F(x) holds. Even that does not give you complete information, since you wouldn't know how to answer questions about surjectivity and bijectivity of F without knowing the intended codomain, and that information isn't there.

So I have wondered what is the best way to explain a function, and what is the best way to define formally what it means for something to be a function?

I bought a copy of "The Handbook of Mathematics" and sought for the definition that it gives, but even though this is a pretty thick book, it didn't have any. I thought that was a little weird.

An analysis textbook explained that a function is given by a "formula". That was weird too, since it appears that mapping x to x+x or to 2x as a function from R to R should mean the same function, despite the difference between those two expressions. (The same textbook also occasionally would assume the choice axiom, which is used to produce functions that have no formulas at all.)

Is there any good answer to the question, what is a function really?