mega...i see..it was a typo then or i read it wrong. But i wonder what your typo equation would look like if it were graphed. That would be interesting. I'm going to graph it using qbasic.


:alright:

EDIT

I graphed it and it has some interesting properties. Is not a full parabola.

1. Y never reaches zero in -X direction only becomes infinitly close to 0
2. Is almost linear from about X=-1 to 0 (Most interesting)
3. Y intercept is very close to .6 >>(.599)

:alright:
Posted on 2002-08-26 17:25:25 by IwasTitan
mega: you are wrong. You treat the f'(x) as it was an absolute value, but it's not. Remember how to diff. the fraction?
Posted on 2002-08-27 04:40:33 by Tomasz Grysztar
I am sorry, i do not know what you mean when you say i treat f'(x) as an absolute value?
Posted on 2002-08-27 08:46:02 by mega
' is not g'(x)/f'(x), but /.
g(x) is /(n+1) in your formula.
Posted on 2002-08-27 10:07:21 by Tomasz Grysztar
mega you wrote this

= ( f(x) )^n+1
------------------
f'(x) * (n+1)

I think what Privalov is saying is it should be this:

= ( f(x) )^n+1
------------------
(n+1)

For the general solution which can not be applied to the first equation:>

(x^2+2)^2.dx


Privalov was showing the division rule for differentian:


f(x) = g(x) / h(x) => f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x)^2)

But i think you guys are just are just on two different tracks in your thinking is all.

Privalov....Mega was integrating and you seem to be differentiating.

:alright:
Posted on 2002-08-27 11:30:43 by IwasTitan
No, I meant that his integrating formula is wrong, he was probably checking it with differentiation (I don't know how else could he invent such formula), but you have to do the mistake I described to get back the function you were trying to integrate.

The correct formula is:
I f'(x)dx = /(n+1)
And the Mega's one is:
I dx = /
but you can not divide the equation by f'(x) because it's not an absolute value (it's x-dependent).
Posted on 2002-08-27 13:58:57 by Tomasz Grysztar
Privalov ..this method that Mega is showing is accepted because that is what i was taught in calculas. What you are saying is that it is not really accurate or is not a general solution because of the x dependence and i agree with you.

Mega is just stating that in the case of the equation below the accepted method of integration is the one he posted.


(X^2+2)^2.dx

( f(x) )^n+1 2 sets of brackets represents the entire equation
------------------
f'(x) * (n+1)


The f'(x) = 2x which is in the bottom part of the equation.

In this case he would be correct unless i'm not looking at it correctly.

I think its just the fact that we all use a different syntax or symbols that is confusing.

But i also must say that i integrated the equation in different 2 ways. First i expanded than integrated then i used the method above. When i plugged in the value 2 into x and calculated the y value they were different. This may be because the value +c is different depending on the method used but i did not check.
edit


Privalov....come to think of it ..from a purely logical view point your absolutly right. You can't take the reciprocal of the derivative of part of the equation and throw it into the integral to satisfy the derivitive of the integral and then expect it to mean anything concrete. But thats what their teaching as stupid as it may seem.



:alright:
Posted on 2002-08-27 15:17:19 by IwasTitan
Originally posted by Privalov
The correct formula is:
I f'(x)dx = /(n+1)
And the Mega's one is:
I dx = /
but you can not divide the equation by f'(x) because it's not an absolute value (it's x-dependent).


I do not understand why you have a problem with dividing by f'(x). If you differentiate the solution i presented, you will get back to the originial problem...

Differentiate with respect to x:
(x^2+2)^3
-------------
6x

will give you:

(x^2+2)^2

Obviously i have missed your meaning somewhere? I understood that you were stating that integrating this eqn was hard to do. I don't think it is. I have told you why?

At the very least, perhaps this will encourage the makers to grant us a maths forum??
Posted on 2002-08-27 19:33:50 by mega
mega,

The reason why Privalov mentioned about the 'absolute value' is that you need to go through 'change of variable' technique using Jacobian. You can find a more detailed explanation in any calculus textbooks.

You are right when you say:
dF(x)/dx = f(x) => F(x) = \int f(x) dx + some const

But, your logic jumped from there. That is, your next claim is:
dF(x)/dx = f(x)g(x)
=> 1/g(x) dF(x)/dx = f(x)
=> F(x)/g(x) = \int f(x) dx + some const.
That is where you fell into the trap.
You have to find out if d(F(x)/g(x))/dx is really f(x). There are special cases in which it happens to be true, but it is not true in general.
Posted on 2002-08-27 19:57:07 by Starless
Mega i did some calculating.


With your method of intigration you end up with this.



(x^2+2)^3
-------------
6x


Now use the division rule for derivitive instead of the chain method.

f(x) = g(x) / h(x) => f'(x) = (g'(x) * h(x) - g(x) * h'(x)) / (h(x)^2)


x^2+2)^3
-------------
6x


dy/dx = 6x*(x^2+2)^2*6x - 6*(x^2+2)^3

____________________________
36x^2



Substitute in the derivitive above x =2 and you will get 27

Now substitute 2 into (x^2+2)^2 and you will get 36

You see privalov is right. The division rule is an accepted method so both equations should give the same result but they do not. I thought your method was taught to us in calculas but i was thinking incorrectly.

You see there is no general method to solve such an equation when X is to a power greater than 1 and enclosed in brackets. It must be expanded to integrate it.

Look closely above and you will see what i say makes sense.

Hey no shame ...i was thinking the exact same way since we started this discussion and even discussing this stuff shows you have some sharp tools in the shed.

At the very least, perhaps this will encourage the makers to grant us a maths forum??

I agree.



:alright: :alright:
Posted on 2002-08-27 20:04:37 by IwasTitan
Originally posted by IwasTitan

At the very least, perhaps this will encourage the makers to grant us a maths forum??

I agree.


I thought the request for the math forum was about numerical method in asm, not about the math itself.
:confused:
Posted on 2002-08-27 20:10:56 by Starless
Starless ...i think they go hand in hand don't they.

The body can not live without the mind.

:alright:
Posted on 2002-08-27 20:20:28 by IwasTitan
Yes, i bow my head in submission, a silly mistake on my behalf...

i will shut up now.
Posted on 2002-08-27 21:01:48 by mega
Mega...don't shut up...your input is respected.

Read back a few posts where i graphed your typo equation.

If Mathematicians were willing to just sit back and rely on present rules we would be no where.

I say break the rules.

We learn from each other what we can.
Posted on 2002-08-27 22:08:32 by IwasTitan
What about numerical methods?
Posted on 2002-08-28 16:45:33 by wolfao
Wolfao ...what are you referring to? If you are referring to my last comment about breaking the rules i'm just saying "experiment".....I'm not saying have no respect for the logic in numerical method and the axioms on which they are based.

I believe in a previouse post you also were referring to my analysis of primes in a non conventional manner as being "games". Whats wrong with experimentation?


:alright:
Posted on 2002-08-28 20:41:09 by IwasTitan
There is a mistake here.
I'm talking about another theme related with the Math Forum.
I would like to know if someone want to discuss numerical methods too.
We could treat the Finite Element Method, for example.

Thanks in advance.
Posted on 2002-08-29 07:40:01 by wolfao
Wolfao said:

We could treat the Finite Element Method, for example.

You are speaking of finite mathematics which encompass combinatorics, permutations, probability, groups, Galois fields,etc etc?

Yah i'm all for that.






:alright:
Posted on 2002-08-29 10:12:11 by IwasTitan
WOLFAO WROTE:

There is a mistake here.


Yes the mistake was in one of my earlier posts in this thread where i hypothesis that.... the sum of the digits of of the difference between the square of a prime and the previouse .....with continuing process untill a digit from 1 to 9 is reached...... is always a multiple of the number 3 for differences that repeat in the sequence.

I had said that all but three of the differences of the suequence i posted were multiples of 3 by the above method. That was a mistake. There are actualy 4.

:alright:


I would have edited the post but it was too late when i discovered it and the time cap on editing had expired.

Hey no baseball strike!!!!!
Posted on 2002-08-30 21:21:04 by IwasTitan
A would like play with The Finite Element Method to solve Partial Differential Equations using Assembly.
Posted on 2002-08-31 07:17:32 by wolfao