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Complex Problems - September 17, 2010
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Spiked Math Comic - Complex Problems vs Real Problems

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Mmm... Bad branch! Get back in your tree!

what the fuck just happened there? im mind=blown from that complex problem. there has to be a problem somewhere!!!

(-1)^.5 * (-1)^.5 = -1
You cannot just put a sqrt above your two numbers like that ;)

I hope you never try quaternions. But what's so hard to understand about the real problem? Spivak, when he gets to the Trig part of his *Calculus* textbook, started by defining π/2 as the integral from -1 to 1 of √(1-x^2)dx. By that definition, the area of the unit circle is trivially 2 x π/2 = π.

I was freaking out for a second, because when I read "area of a circle," my dumb first thought was "A=2(pi)r" and saw (pi) as the answer instead of 2(pi).

Oh, what a fool I am.

that works because every positive number has a negative and a positive square root, right?

also check out my new gravatar, I drew it myself

Nice--my gavatar seems broken on this site...


Now it's working again. :^)

Ya, I no idea what happened. I didn't touch the code or anything.

Sometimes things (like computers) just need attention. That's the reason, I think, for the "Check Engine" light in your car: it comes on when the engine needs attention. So pull over, pop the hood, check to be sure the engine is still there and still looks like an engine, then get back in and drive.

I guess it comes and it goes--but mostly it goes.

pi isn´t an angle anymore.... :P!!

The problem with the complex numbers proof is that x^2 is not invertible, so sqrt is only a function if you restrict its codomain appropriately. Keeping this in mind f:{xi:x is positive}-->{x:x is negative} f(x)=x^2 has inverse f^-1:{x:x is negative}-->{xi:x is positive} g(x)=sqrt(x). The problem is that (-1)(-1)=1 is not in the domain of g.

Sorry, by g i meant f^-1.

Hey! I sent you this a long time ago and you didn't even mention it! But I forgive you ;)

"Hi Jasper,

Thanks for the message. Love the 1 = -1 argument, and it would
definitely fool a lot of people :D

All the best,

The 'dt' is neither infinite small nor orthogonal to the circle :-P

do I miss something or pi*r*r = pi*1*1 = pi?

Yes, you are missing the fact that the formula you used comes from the process depicted in the comic. To find pi*r*r you need only to substitute the integral interval to 0 to r.

It only goes wrong at the square root of 1 - which can be plus or minus one, but in this case only one of those choices is correct!

Square root is an operation, and, as such, is a function. That means, there can be only one value of square root.
If you ask how many real values are 1 when squared, then you have two values

obs: Highlander quote!

That's an interesting way of calculating that area. I generally prefer the double integral method myself.

so it's \int_0^{2\pi} d\phi \int_0^1 t dt :).

Actually, I usually write it with the integrals together, so:
\int_0^{2\pi} \int_0^1 r dr d\phi

My new favorite useless piece of information:

Wikipedia has a full page of calculations proving that 22/7 exceeds π.

I would've just checked with my calculator.

sqrt(a) * sqrt(b) = sqrt(ab) iff a ≥ 0 or b ≥ 0

Tricky error in the top example Mike. When combining surds you have to take the negative terms out and convert them to i's.
Then you'll get:     i.i.sqrt(1).sqrt(1)

BTW one of the first "proofs" I learned at university was 1 equals 2 :)

Just a bit busy atm. Will start posting regularly soon.

I always use geometry to solve geometry problems. I just think its appropriate.

The area of a circle is the area of a square minus the area of four "half" cones on each vertex. Using the radius we can calculate the following


( 4r ) - { 4 [ &pi r ( &radic 2*2 + h*h ) /2 ] }

Contrarily, the complex problem failed on the "real" part.

The way my geometry teacher taught it was to imagine cutting a circle into slices and fitting them together. Then as you cut thinner and thinner slices, it begins to tend to a rectangle. From there, you can easily find the area to be pi*r^2

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