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"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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Where in the world did you get that?
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Where in the world did you get that?
What exactly? What letters or signs cause in you misunderstanding?
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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Explain how you got the first equation...
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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Explain how you got the first equation...
Look this:
h ttp://vladimir938.eto-ya.com/files/2012/12/screenshot-14.12.jpg
point 3.
Last edited by 21122012 (2012-12-15 07:36:08)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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seems to me that. And you as think?
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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seems like:
waaaaa
30+2=28 (Mom's identity)
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hi Fistfiz
If you accept the premise that 1 = 0, then you don't need calculus to get 2 = 1 (just add 1 to each side).
Alternatively suspect that 1 isn't 0 after all.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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hi Fistfiz
If you accept the premise that 1 = 0, then you don't need calculus to get 2 = 1 (just add 1 to each side).
Alternatively suspect that 1 isn't 0 after all.
Bob
cool!
30+2=28 (Mom's identity)
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seems like:
waaaaa
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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I think Mathematical analysis says that
.21122012 wrote:I think Mathematical analysis says that
.
How can you divide by zero? May be you wanted to write:
Hi round;
What does that mean??
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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Here help:
:)"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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Hi;
So 0 = 1? Gave me an idea!
Here is what I did. I had a 1 dollar bill in my pocket so I took it out and placed it in the drawer. But since I now had 0 dollars in my pocket immediately a dollar appeared in there. After all 0 dollars = 1 dollar. So I thought why not try this again. I took the dollar out and immediately it reappeared in my pocket! This worked again and again... Now the question is when should I stop?
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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It too difficult for me. I have not 1 dollar
Last edited by 21122012 (2012-12-16 13:30:18)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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Here is what I did. I had a 1 dollar bill in my pocket so I took it out and placed in the drawer. But since I now had 0 dollars in my pocket immediately a dollar appeared in there. After all 0 dollars = 1 dollar. So I thought why not try this again. I took the dollar out and immediately it reappeared in my pocket! This worked again and again... Now the question is when should I stop?
Stop when you have enough to live comfortably + 1 dollar. Send that 1 dollar to me. I have large pockets.
hhmmm. I've seen this before.
Try
So
That looks OK to me.
Sorry bobbym Your chance to become richer than Bill Gates has been dashed.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob
The problem with that is that the constand of integration appears after integration... In his steps, he never actually differentiated...
The flawed step is assuming that -Integral[1/x,x]+Integral[1/x,x]=0 instead of Integral [0,x].
Last edited by anonimnystefy (2012-12-16 22:14:28)
Here lies the reader who will never open this book. He is forever dead.
Taking a new step, uttering a new word, is what people fear most. ― Fyodor Dostoyevsky, Crime and Punishment
The knowledge of some things as a function of age is a delta function.
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hi Stefy,
You may be right. My point is that this proof starts with a formula that I call 'integration by parts'. This formula is obtained from the product rule, so differentiation is involved.
It is commonly assumed that integration is the reverse of differentiation. I've even met teachers who teach this and books that state it. It isn't true.
Differentation is a prescriptive process for working out a gradient function, say f(x), from a function, say F(x).
Integration is a summation process. That's where the integral symbol comes from. It is a stylised S.
http://en.wikipedia.org/wiki/Fundamenta … f_calculus
This makes clear that, if you have obtained F(x) by integration, then working out dF/dx gives f(x) once more. But starting with F(x) and differentiating cannot be reversed to get F(x) back as there are infinitely many antiderivatives.
So whenever you use the antiderivative to obtain an integral, you must insert a constant of integration and failure to do so may well result in an answer that is wrong by a numeric amount.
I showed this at
http://www.mathisfunforum.com/viewtopic.php?id=18614
I consider the original statement to be correct, save for a numeric amount.
In my opinion, that's why the proof appears to show an incorrect result, 1 = 0
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi Bob,
if I may, it seems to me that the (logical) error is deeper:
because
30+2=28 (Mom's identity)
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hi Fistfiz,
Have you looked at
http://www.mathisfunforum.com/viewtopic.php?id=18422
There's a lot of it, so I don't expect you to look at it all. Post #1 is worth a look though.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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I had a quick glance and it is very obscure to me; i'll try to read it later, thank you.
30+2=28 (Mom's identity)
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So
That looks OK to me.
Sorry bobbym Your chance to become richer than Bill Gates has been dashed.
Bob
Give to me the website where this miracle except as here is still written?
This incorrect equality!
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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Hi Bob
The problem with that is that the constand of integration appears after integration... In his steps, he never actually differentiated...
Last edited by 21122012 (2012-12-17 06:17:47)
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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Hi Bob
The flawed step is assuming that -Integral[1/x,x]+Integral[1/x,x]=0 instead of Integral [0,x].
You are wrong. Such formula isn't present. Here a problem in other! Here a problem in an error of Calculus!
"The conditions imposed on functions, become a source of difficulties which will manage to be avoided only by means of new researches about the principles of integral calculus"
Thomas Ioannes Stiltes. ... I made it!
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