# MandrakeRoot

Senior Members

194

## Posts posted by MandrakeRoot

### Hypergeometric formula

You still didnt say what is x. But your algorithm is exponential in the size of your problem instance D. (since you have a lot of terms of the type x^(something with D), like D, D - 1 etc...

The bigger D, the more terms you have. The only reason in the last term X^D, you have something like x^D/D! is because you have D! x^n/n!^2 for n =D, so maybe you can estimate the terms before ?

Something like the coefficient of the n-1ste term behave at least like .... (something with n) and at most like (something with n), that way you know how your function behave approximately and you can send n to infinity and maybe use some of the assymptotic formula for hypergeometric fucntions. (Remember that they are each a solution to some differential equation).

Mandrake

### Hypergeometric formula

What is x then ?

Does it depend on D is some way ?

(If x(D) = D! then your above limit argument is flawed).

Mandrake

### Hypergeometric formula

What is D ?

If D > 0 and if D is an integer indeed the series will terminate and your hypergeometric function will be a polynomial.

What do you mean with x^D/D! converges to zero ? (You should at least keep x fixed in such an argument).

Anyway the way you write down the hypergeometric function the D! will be in the upperpart of the fraction, so you have more something like D! x^D then x^D/D!

A quick maple calculation shows that hypergeom(-D;2:-x) behaves like :

1 + 1/2 D x + 1/12 D (D - 1) x^2 + 1/144 D (D - 1) (D - 2) x^3 + 1/2880 D (D - 1) (D - 2) (D - 3) x^4 + 1/86400 D (D - 1) (D - 2) (D - 3) (D - 4) x^5 + 1/3628800 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) x^6 + 1/203212800 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) (D - 6) x^7 + 1/14631321600 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) (D - 6) (D - 7)x^8 + 1/1316818944000 D (D - 1) (D - 2) (D - 3) (D - 4) (D - 5) (D - 6) (D - 7) (D - 8) x ^9 + O(x^10)

Your algorithm looks more factorial than anything else i think.

Mandrake

### Differentials and Integrals(uggh!)

Yes indeed, the notation is rather sloppy. It would be better to write y(X), y'(X) and Y(X) for example (and indicating the domain and range of these functions !)

Mandrake

### Differentials and Integrals(uggh!)

I think with integral you mean primitive here !

Mandrake

### Partial Derivatives Help

Can't you use the binomial formula or something like that ?

If you can do it at (0,0), i think you are surely pretty close to showing it for every couple (x,y)

Mandrake

### x not= x

Doron claims that Dedekind cuts do not work because the limit point of a sequence (e.g. sequence of rationals) is always separate by a finite interval from its "limit" irrational (and due to the self-similar scaling of the real line' date=' any gap is a "1" gap.

Here's his latest set of posts:

Doron always talks a lot, i wouldnt pay attention to his pseudo-intuitive arguments.

Mandrake

### x not= x

Yeah that is pretty much a typical discussion with Doron :=)

Mandrake

### x not= x

But you can't verify applied mathematics, since x not= x.

You still dont see the difference between application and theory. It is hopeless.

Mandrake

### x not= x

Mathematics is not an empirical science. It is based on logical deduction. When doing applied mathematics it is important to verify your results with observations.

It is true that physical laws have restrictions on their domain of use. In pure mathematics restrictions are clearly formulated in the theorems and lemmes. When applying math to the "real world" (whatever you may mean with that) it is up to you to verify that these conditions are satisfied. (Like in your example above that indeed all scientists are aliens). The fact that not all scientists are aliens does not in any case make untrue the conclusion "Einstein is a scientist => Einstein is an alien", under the hypothesis that all scientists are aliens.

Mandrake

### proving the theorems of limits of functions

Some hints :

1) Uniqueness is quite trivial (Suppose there are two functions that satisfy the conditions, look at their difference and use the fact that for every point x in the closure of U there is a sequence entirely in U that converges to x, to show that both functions have to be equal. (Here you need the fact that the difference of two contniuous functions is again a continuous function)

2) => Take any point x in the closure of U and use the above fact about the sequence entirely in U that converges to x. Use uniformy continuity to conclude that f(x_n) is a Cauchy sequence in Y => hence convergent.

Define g(x) to be this limit.

3) Check that g is well defined and does not for instance depend on the choice of the sequence x_n above

4) Show that g is the same as f on U

5) Show that g is continuous.

Mandrake

### proving the theorems of limits of functions

yes please. some hints would be good.

could u give me some cauchy sequence which does not converge.

 oops x_n=a for all n is a cauchy sequence that doesnt converge rite[/edit] where d is the usual metric in R

 dOH.' date=',,, that ovbisouly converges

it should be x_n = n[/quote']

No that is not a Caucy sequence. X_n = n does not stick together in the tail. R is a complete metric space so convergent sequence and Cauchy seqeunce are the same notion here. The notion of Cauchy sequence is a sequence where at the end of the sequence all elements are arbitrarely close on to the other.

Mandrake

### proving the theorems of limits of functions

Yes i could. Consider the space (Q,d), where Q is the set of fractions and d is absolute distance (like the metric on R in fact) (note that we will pretend that R does not exist and therefore in the above replace epsilon by 1/k)

Take the following sequence defined by $q_{n+1} = q_n - \frac{q_n^2 - 2}{2q_n}$ with q_0 say 2;

This is clearly a sequence in Q that should converge to sqrt(2), but well that is not an element of Q.

Mandrake

### proving the theorems of limits of functions

A complete metric space is a metric space, but in this metric space all Cauchy-sequences are convergent. Note that in any metric space all convergent sequences are Cauchy !

(A Cauchy sequence is a sequences that sticks together in the tail :

Let {x_n}_n be a sequence in the metric space (X,d), then {x_n}_n is called a Cauchy sequence if for every epsilon > 0 there exists a N (integer) such that

d(x_n,x_m) < epsilon for all n,m >= N)

Often it is easily shown that a sequences is Cauchy and so using the fact that your space is complete you know that there is some limit for your sequence !

If you would like some hints for the proof i could write them down ?

Mandrake

Ps: it is not a silly question

### x not= x

Do you realise that you just opened the door to insanity? If it does not have to be consistent with the real world' date=' then anything goes! All scientists are aliens. Einstein was a scientist. Therefore, Einstein was an alien!

[/quote']

See you seem to get basic logic after all !

This is a logically correct statement with little pratical use.

Practice is the touchstone of every theory.

No it isnt. When einstein wrote down his theory most of the consequences could not be checked. It was not until much later that some empirical data could be obtained to demonstrate the "applicability" of his theory. Basically you are saying that if we are a primitive culture and cant check let's say Newtons laws' date=' newtons laws are not a proper theory.

The same instant? Pardon me sir, but how long is an instant exactly? Is it an infinitesimal interval of time? But then the apple will be subject to minute changes. Or is it no time at all? A zero of time - what a meaningless fancy. Everything exists in time; time is consequently a fundamental element of existence. The axiom x = x signifies that an object is equal to itself if it does not change, that is, if it does not exist.

That's right - an instant does not exist. It is just another figment of your mind. I shall commend you on your very lively imagination. But imagination only makes good science fiction, not good science. What we have here is a prime example of philosophical idealism - you take the abstraction, which is after all only an approximation of the real world (i hope you don't disagree with that), and hand it back to the real world, expecting the real world to follow your dictat. It is a terrible shame that so many scientists and science-interested people are ignorant or contemptuous of philosophy. So many - there is no milder expression - so many infantile mistakes could be avoided!

Yes indeed an infinitisemal instant of time. Many physical formula are derived using this notion, and like you seem to say "if it works it is true", since these formula seem to be supported by empirical data, the abstraction of using "infinitisemal instants" is a not too absurd one.

Mandrake

### proving the theorems of limits of functions

Let me write down the entire theorem

Let $(X,d)$ be a metric space and $(Y,\rho)$ a complete metric space. Let $f : U \rightarrow Y$ be a uniformly continuous function on the subset $U \subseteq X$ of X. There exists a unique continuous function $g : \bar{U} \rightarrow Y$ such that $g|_{U} =f$.

The proof is quite simple really, i am sure you can do it bloodhound.

Mandrake

### proving the theorems of limits of functions

Well the real topological definition of a compact set is a little more complicated. The equivalence closed and bounded and compact only holds in specific topological spaces (like the R^d's). In topological vector spaces compact implies bounded and closed but the inverse is not perse true.

A closed set that is not bounded in R would be [0,infty)

If i would give you a uniformly continuous function on some open interval, with the above argument it could be immediately extended to a continuous function on the closed interval that is the closure of this open interval, so i dont see how we could give non-trivial exemples ?

In fact the whole strenght of the theorem is when we consider functions on some metric space X (not even supposed of finite dimension), then the closure of a set is less trivial and also the extension to go with. Or functions on some general open sets of R (that could be the union of many open intervals)

Mandrake

### proving the theorems of limits of functions

I will give an example. If f :(a,b) -> R is some uniform continuous function, then it is possible to find a continuous function g: [a,b] (the closure in R of (a,b)) -> R such that f(x) = g(x) for all x in (a,b), this g would be the extension of f to the closure of (a,b)....

Do you see what i mean ?

Mandrake

### x not= x

"In first grade books' date=' integers are depicted as fruit. An apple plus an apple equals two apples. This is a useful concretisation. If mathematics is correct, then this kind of concretised example must be correct, since "the proof of the pudding is in the eating" - the proof of the theory is in its applicability to the real world."

[/quote']

No : Mathematical statements are logically derived from the axiomatic system they are statements in (ZFK if you like) and are therefore independant from any "real-world" notion whatever that may mean.

The saying "the proof of the pudding..." is really the most ridiculous thing i have ever heard !

Your statement is something along the following line : "I roll a dice and roll 6, so i conclude every dice roll will always result in a 6."

Even if an apple changes over time, in what way does that make the apple different from himself ? If you look at two different instances of time, yes the apple has evolved and will no longer be in the same state, but when looking at the same time instance it will be the same thing.

Mandrake

### Applications of the Gamma function

How about the fact that the Gamma function can help in calculating integrals and simplifying expressions. Not to forget that you can use them to make a short hand notation for the pochhammer symbols. The gamma function often comes about in special function theory and is therefore also usefull in finding analytic solutions of some DE's !

Mandrake

### proving the theorems of limits of functions

Maybe he wants to know the following. If f is a uniformly continuous from some set A, a subset of a metric space (lets say a subset of R) into some metric space (lets say R again), then f has a continuous extension on the closure of A into R (or the appropriate space).

There are quite a lot of different theorems on limits though.

I doubt he wants you to check that the space of continours functions from some vectorspace X into some vectorspace Y is a vectorspace.

Mandrake

### x not= x

I'd rather have "matt grime 1 - doron shadmi {}"

Haha this one is funny. Though i think doron would say 1_1 or 1_0_1

Mandrake

### x not= x

I knew there was something about why all this about xnot=x didn't fit...I think you got it there' date=' yes.

I'd like to see someone say that you cannot claim that 2 apples are never two apples, no matter which appels you've got.

Another example. Say you have 3 chairs : one arem chir, one deckchair and one stool. Youc an say swap the stool for a dinning chair. The stool certainly is different from the dinning chair, but you still have 3 chairs.[/quote']

I think you are missing the whole point of abstraction here. The whole point of applying the abstract consistent number system is that you can quantify something. Depending on what is your question the number system can help you quantify. If you ask for three apples, then someone can give you red or green ones. If you specify that you want red apples,...well you get the point. But that doenst have any impact on the number system. If you consider that a red apple is not the same as a green one, then you cant call it both "x", i.e., it is not the same variable.

The whole point of talking in terms of apples/parts of a pie or whatever is to illustrate abstract mathematical concepts and not define them !

Since it is a little hard to start in kindergarten with real mathematics, it is done more intuitively.

Mandrake

### x not= x

I think you should separate abstraction and application. You can perfectly create the natural numbers from a simple axiom using set theoretical arguments. Now when you would like to apply this system you decide that the unity is an apple regardless of its size, colour, configuration, weight etc...... If you disagree with this decision and find that ambiguous and consider that a red apple is not the same as a green one, then that in no way changes the abstract system. The only comment you are making is on the application of the abstract system to the real world.

Mandrake

### Help Solve a Group of Equations

Yes! It's right.

Thanks Mandrake' date=' I never know where to start with more than 2 equations...[/quote']

Great you got it. It is true that several non-linear equations can be quite annoying to solve;

With a unique solution i mean that there is only one couple (a,b,c,d) that satisfies your system of equations. Taking for instance :

a*b = 10, surely has more then one solution as a system.

As does a^2 -1 = 0, though it is one equation with only one unknown. I dont know where you got this system, but if it is from some physical reasoning, you might know intiutively that there is only one solution.

Mandrake

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