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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

**"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.** ...

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Where in the world did you get that?

Here lies the reader who will never open this book. He is forever dead.

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

anonimnystefy wrote:

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.** ...

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

Explain how you got the first equation...

Here lies the reader who will never open this book. He is forever dead.

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

anonimnystefy wrote:

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.** ...

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

21122012 wrote:

seems to me that. And you as think?

** Thomas Ioannes Stiltes.** ...

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

seems like:

waaaaa

30+2=28 (Mom's identity)

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

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

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

bob bundy wrote:

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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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

Fistfiz wrote:

seems like:

waaaaa

** Thomas Ioannes Stiltes.** ...

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**Pshk****Guest**

21122012 wrote:

I think Mathematical analysis says that

.**round****Guest**

Pshk wrote:

21122012 wrote:I think Mathematical analysis says that

.

How can you divide by zero? May be you wanted to write:

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

Hi round;

What does that mean??

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

21122012 wrote:

Here help:

:)** Thomas Ioannes Stiltes.** ...

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 83,251

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?

I have the result, but I do not yet know how to get it.

All physicists, and a good many quite respectable mathematicians are contemptuous about proof.

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

It too difficult for me. I have not 1 dollar

*Last edited by 21122012 (2012-12-16 13:30:18)*

** Thomas Ioannes Stiltes.** ...

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

bobbym wrote:

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.

21122012 wrote:

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

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**anonimnystefy****Real Member**- From: The Foundation
- Registered: 2011-05-23
- Posts: 14,886

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.

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

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

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

Hi Bob,

if I may, it seems to me that the (logical) error is deeper:

because

is just a symbol to denote the class of antiderivatives; so, saying class=number makes me think 21122012 is totally missing the meaning of it all.

30+2=28 (Mom's identity)

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**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,269

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

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**Fistfiz****Member**- Registered: 2012-07-20
- Posts: 33

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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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

bob bundy wrote:

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!

** Thomas Ioannes Stiltes.** ...

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

anonimnystefy wrote:

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)*

** Thomas Ioannes Stiltes.** ...

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**21122012****Member**- Registered: 2012-11-16
- Posts: 278

anonimnystefy wrote:

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!

** Thomas Ioannes Stiltes.** ...

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