1 = 0

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gogobera
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1 = 0

Postby gogobera » Tue Nov 21, 2006 12:47 am UTC

I didn't make a solution thread when I should've, but now I have. It's here: http://forums.xkcd.com/viewtopic.php?t=660

I've always enjoyed false proofs, but the only one I saw on here was an algebraic one. Here's one I stubled upon way back when I was learning the Calculus (do people mind TeX notation? It shouldn't be too cryptic.)

To evaluate the integral of 1/x, we use (the much loved) integration by parts. Recall that integration by parts exploits the following:

\int u dv = uv - \int v du.

Let u = 1/x, then du = -1/x^2 dx. Let v = x; therefore, dv = dx. We write
\int 1/x dx = \int u dv
\int 1/x dx = uv - \int v du
\int 1/x dx = (1/x)x - \int x (-1/x^2) dx
\int 1/x dx = 1 + \int 1/x dx.

Subtracting the original integral from both sides leaves 0 = 1. QED Note, this solves the pesky problem of having to ever evaluate (or even define) the log base e (an irrationally based logarithim, if ever one was!).

Anyone seen that before? I haven't.
Last edited by gogobera on Tue Nov 21, 2006 1:51 pm UTC, edited 1 time in total.

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svk1325
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Postby svk1325 » Tue Nov 21, 2006 4:43 am UTC

I get the feeling that this has something to do with 1/x and -1/x^2 being discontinuous at x=0.
"Insanity in a measured dose is a good thing - the difficulty lies in the measurement."

Erasmus
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Postby Erasmus » Tue Nov 21, 2006 5:52 am UTC

When you evaluate an indefinite integral, ... there's an arbitrary constant:

\int 1/x dx
= \int u dv
= uv - \int v du + C
= x (1/x) - \int x (-1/x^2) dx + C
= 1 + \int 1/x dx + C

Therefore
0 = 1 + C

Therefore C = -1.

A "reverse" way to write the same "proof" is to start with u and v and evaluate d(uv)/dx using the product rule (which is the inverse of integration by parts):

u = 1/x => du = -1/x^2 dx
v = x => dv = dx
d(uv) = u dv + v du = dx/x - x dx/x^2 = 0

Your proof at this point involves integrating both sides to get uv = \int u dv + \int v du, but when you integrate, the arbitrary constant has to come in: it's really uv = \int u dv + \int v du + C.

One last way of looking at it: do a definite instead of indefinite integral:

\int_a^b 1/x dx
= [ x/x ]_a^b + \int_a^b 1/x dx (a, b > 0)

Doing the same subtract-from-both-sides we get
0 = [ 1 ]_a^b
which is quite obviously the case.

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SpitValve
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Postby SpitValve » Tue Nov 21, 2006 7:25 am UTC

yay! maths is saved!

gogobera
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Postby gogobera » Tue Nov 21, 2006 1:49 pm UTC

Seems I oughtta have created a "solution thread" ... hmm. oops. Ok, doing that now....

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Pathway
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Postby Pathway » Fri Nov 24, 2006 3:57 am UTC

Too late. It's been quite convincingly solved.

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Minerva
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Postby Minerva » Thu Nov 30, 2006 2:28 pm UTC

Can anyone tell me why the well known result for this problem - Ln(x) + C - doesn't fall out of the solution, using the simple integration by parts as above?

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Cosmologicon
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Postby Cosmologicon » Fri Dec 01, 2006 1:18 am UTC

Because integration by parts isn't the way to solve this one?

Generally speaking, integration by parts just produces another true statement. Sometimes the integral in that statement is simpler and you can solve the integral, but not always. Another great example is the integral of exp(x)sin(x)dx. You can solve it using integration by parts twice even though the integral you get out of it is no easier to evaluate!


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