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Old May13-09, 11:19 PM       Last edited by protonchain; May13-09 at 11:34 PM..            #1
protonchain

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Why is this proof wrong

Another one of those 0 = 1 proofs.

If you dislike them, please navigate away to cats doing things with captions. Otherwise stay tuned.

Tell me what's wrong (obviously I know what's wrong but I just want to put it out there for people to mull over).

Note, you need to know basic calculus.

LaTeX Code: <BR><BR>\\int \\frac{1}{x} dx = \\int \\frac{1}{x} dx<BR>

LaTeX Code: <BR>u = \\frac{1}{x}<BR>

LaTeX Code: <BR>dv = dx<BR>

LaTeX Code: <BR>du = \\frac{-1}{x^2} dx<BR>

LaTeX Code: <BR>v = x<BR>

LaTeX Code: <BR>\\int \\frac{1}{x} dx = u * v - \\int v du<BR>

LaTeX Code: <BR>\\int \\frac{1}{x} dx = \\frac{1}{x} * x - \\int x * \\frac{-1}{x^2} dx<BR>

LaTeX Code: <BR>\\int \\frac{1}{x} dx = 1 - \\int \\frac{-1}{x} dx<BR>

LaTeX Code: <BR>\\int \\frac{1}{x} dx = 1 + \\int \\frac{1}{x} dx<BR>

LaTeX Code: <BR>0 = 1<BR>
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Old May13-09, 11:28 PM                  #2
Pengwuino
 
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Re: Why is this proof wrong

Right off the bat, LaTeX Code: u = \\frac{{dv}}{x} = dx makes no sense. The second line makes no sense either.... am i completely missing something?
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Old May13-09, 11:32 PM                  #3
protonchain

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Re: Why is this proof wrong

Sorry those are supposed to be separate. I will edit that in.

This is what it should look like

LaTeX Code: <BR>u = \\frac{1}{x}<BR>

LaTeX Code: <BR>dv = dx<BR>

I have also added an extra step just to show that I am going to be using integration by parts to do the "proof"
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Old May14-09, 12:25 AM                  #4
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Re: Why is this proof wrong

LaTeX Code: \\int \\frac{1}{x} dx = 1 - \\int \\frac{-1}{x} dx isn't valid. Integration by part goes like:

LaTeX Code: \\int\\limits_a^b {udv}  = [uv]_a^b  - \\int\\limits_a^b {vdu}

The term you think is 1 is actually 0.
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Old May14-09, 03:51 AM                  #5
Russell Berty

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Re: Why is this proof wrong

We are dealing with antiderivatives. So, given F(x) and G(x) that are antiderivatives of 1/x, it is true that F(x) = G(x) + C, for some constant C. For example, the second to last line you have can be written as

ln(x) + C = 1 + ln(x) + D for some constants C, and D. We do not then conclude that 0 = 1.

The integral of 1/x dx is a FAMILY of functions that all differ by a constant.
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Old May14-09, 05:04 AM                  #6
neu

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Re: Why is this proof wrong

I only got to the end of the second sentence
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Old May14-09, 05:44 AM                  #7
D H

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Re: Why is this proof wrong

Russell has the right answer. To be strictly correct, the rule for integration by parts is better written as

LaTeX Code: \\int u dv = u*v - \\int v du + C

We drop the arbitrary constant because it is implied by the very use of indefinite integrals, or antiderivatives. The inverse of the derivative is not unique.
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Old May14-09, 05:46 AM                  #8
protonchain

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Re: Why is this proof wrong

Russell and D_H are correct, the constant term is missing. Gj guys :)
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Old May16-09, 04:16 PM                  #9
gel
 
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Re: Why is this proof wrong

Originally Posted by protonchain View Post
Russell and D_H are correct, the constant term is missing. Gj guys :)
and Pengwuino too. You either put in constants of integration, or use definite integrals. Either way resolves the error.
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Old May16-09, 04:38 PM                  #10
protonchain

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Re: Why is this proof wrong

Right, but since I defined the problem in the start as indefinite integrals, I was looking for the answer that was related to indefinite integrals. Anyways. It's just a simple problem
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