I tried to look at szk_kei's proof, but when I went to his site and clicked the link to it I just got an acrobat reader thing with 4 blank pages. Probably my computer doesn't support something that it needs to. I tried it out though, and it seems to work.

y = x

∫ydx = 0.5x² + c
∫xdy = 0.5y² + c

∫ydx + ∫xdy = 0.5x² + 0.5y² + c

x = y ∴ x² = y² ∴ 0.5x² + 0.5y² + c = x² + c
x = y ∴ x² + c = xy + c

Last edited by mathsyperson (2005-11-08 07:58:09)

y=f(x):differentiable

(xy)'=y+x(dy/dx)
xy+C=∫ydx + ∫x(dy/dx)dx
        =∫y dx + ∫x dy

If you want to sea its detail, please visit my site http://www.freewebs.com/keiichi_suzuki/
->[The Study (Mathematics created by myself)]

When I show an expression without proof, they say always "It can never hold".
When I show an expression with proof, someone says "It is only an application of integration by parts",
someone says "It is simple and beautiful".

Your definition is monotone increasing function passing through the origin and  differentiable.
My definition is only differentiable.

Yes, it was. I'll do the y = x², because it gives me a chance to practice the new code things.

y = x²
∫ydx = x³/3 + c

x=√y
∫xdy = 2/3√(y³) + d

x³/3 + c + 2/3√(y³) + d = x³/3 + 2/3x³ + c + d = x³ + c + d

x³ = x*x² = xy + c + d

c and d are both arbitrary constants, so they can be combined into an arbitrary constant C.

∫xdy + ∫ydx = xy + C. It still works!

Last edited by mathsyperson (2005-12-03 03:49:16)

Hi Mr.Keiichi Suzuki,
Welcome to the forum.
Your website is cool, your proof is elgeant and precise.
I do have  great respect for Japanese Mathematicians, I had read about Shimura and Taniyama before.

I'm not a mathematician either.
I work taxing manufacturers (collecting taxes).

fortunately for me, I'm still a student. : )

Can you see the full explanation of integral relation and property of laurent expansion with an acrobat reader.

http://www.freewebs.com/keiichi_suzuki/

Looks good, szk_kei, though I haven't checked if it is right or not.

And we can form something like that for definite integrals:

Last edited by krassi_holmz (2005-12-29 02:25:27)

>Looks good, szk_kei, though I haven't checked if it is right or not.

Thank you