That One Thread

[Altera the Hun]

Why would I be angry about it anyway? Didn't even notice the undertones the first time I read :o.

and see, I was totally right.

Well, the undertones were the main reason, but...

That's beside the point! Stop changing the subject!

[Thor Odinson]

*Spews Sprite across the monitor* WAIT, WHAT.

The amount of manly I accumulate is over 9000.

I wield Steel Blades with no AS loss.

[Kriemhild]

btw Luminescent Blade, there WAS no dividing by zero actually. The square root of zero is still zero and not "undefined" because there is no other option. This other guy DID point out the error for me though, which actually had nothing to do with zero:

cos²x = 1 - sin²x

=> cosx = (1 - sin²x)^0.5

Only the positive square root of the left-hand side will equal to the positive square root of the right-hand side, while the negative square roots will also be equal to each other. cospi is negative, so it can't be equal to the positive (1 - sin²pi)^0.5) = 1. If there was just a negative sign at the start of either the left or the right, then it would have been okay.

Edited December 8, 2010 by Proto

[Narga_Rocks]

Lolwut.

There was probably dividing by zero involved, since sinπ is 0.

(Sin²x)^0.5

=√(Sin²x)

=(Sin²x)/(sinx)

Let x=π

Sinπ=0

I'm more worried about the part where there was a square root on both sides without considering +/-. That might also be an issue.

AT least, considering how (1 - cos²x) = 2 when x is pi. And 2² = 4. And the right side was 4. And 4 = 4 is not a problem.

You didn't even need to go to the end to have a problem.

cosx = (1 - sin²x)0.5

This line already doesn't work if x = pi. -1 = 1. But of course, if you consider how (-cosx)² = cos²x as well, it's not an issue.

You don't even need to involve cos and sin to make something like this happen. A simple use of square roots while taking the +/- that doesn't work will already make an issue. And any problems that involve the quadratic formula but must always have a positive answer have a similar problem if you take the unfavourable side of +/-.

[El Rey León]

Because you watched Code Geass, which makes you awesome!

[Kriemhild]

I'm more worried about the part where there was a square root on both sides without considering +/-. That might also be an issue.

AT least, considering how (1 - cos²x) = 2 when x is pi. And 2² = 4. And the right side was 4. And 4 = 4 is not a problem.

You didn't even need to go to the end to have a problem.

cosx = (1 - sin²x)0.5

This line already doesn't work if x = pi. -1 = 1. But of course, if you consider how (-cosx)² = cos²x as well, it's not an issue.

Congratulations, Narga! You figured it out before Luminescent Blade did!

You don't even need to involve cos and sin to make something like this happen. A simple use of square roots while taking the +/- that doesn't work will already make an issue. And any problems that involve the quadratic formula but must always have a positive answer have a similar problem if you take the unfavourable side of +/-.

I'm pretty sure this was all done deliberately and using a simple quadratic equation would make the error too easy to detect. Trigonometric identities make it look more complicated, so I guess most people would overlook the square root mistake. And the fact that x = pi was only used at the end, by which time most viewers would have assumed the steps to be correct.

Edited December 8, 2010 by Proto

[Thor Odinson]

LOL Proto and Narga posting at the same time.

But yeah, I see it now.

Anyway, I'm still staring at the other one, going over each step.

x=-1-1/x

@ x=1

1=-1-1/1

1=-2

That's where the issues starts.

The solution to x^2+x+1=0 is imaginary, actually.

It's dividing by imaginary numbers, in a sense.

[Kriemhild]

LOL Proto and Narga posting at the same time.

But yeah, I see it now.

Anyway, I'm still staring at the other one, going over each step.

x=-1-1/x

@ x=1

1=-1-1/1

1=-2

That's where the issues starts.

The solution to x^2+x+1=0 is imaginary, actually.

It's dividing by imaginary numbers, in a sense.

We all know that x² + x + 1 = 0 is supposed to have no real solution. (Even someone who just learned the quadratic equation would probably try putting a, b, and c all equal to 1 since that's the simplest example). Despite this, I still can't find any flaw with that guy's methods. It seems perfectly reasonable but there has to be a mistake SOMEWHERE.

[Altera the Hun]

Just for Proto:

[Kriemhild]

Just for Proto:

Safe?

I'm in a university lab, so my definition of "safe" is a lot stricter now (as if it wasn't strict enough already).

Edited December 8, 2010 by Proto

[Thor Odinson]

I think the flaw IS that x cannot possibly be equal to 1 with that, since he started out with an equation that already have a set solution. So when 1 is inserted, crap happens.

[El Rey León]

That part was sad. :(

[Thor Odinson]

Poor Euphie. ;~;

[Kriemhild]

Just for Proto:

Saddest part of the entire series imo. And Euphie was my favorite character too...

I think the flaw IS that x cannot possibly be equal to 1 with that, since he started out with an equation that already have a set solution. So when 1 is inserted, crap happens.

But WHY can't it equal to 1? Aside from the obvious "x = 1 doesn't solve the equation", how was his/her method that led to x = 1 incorrect? He didn't divide by 0, or mix up positive and negative square roots. And yet, he arrived at x = 1 which is obviously not true.

That part was sad. :(

Indeed

Poor Euphie. ;~;

Zero is evil

[Altera the Hun]

...I... I can't help but laugh...

[Thor Odinson]

But WHY can't it equal to 1? Aside from the obvious "x = 1 doesn't solve the equation", how was his/her method that led to x = 1 incorrect? He didn't divide by 0, or mix up positive and negative square roots. And yet, he arrived at x = 1 which is obviously not true.

I can't see the flaw in the steps either, even after staring long and hard at it. The only possible reason could be that the x=-1-1/x still holds the same solution as the original equation and thus plugging in any value for x other than the imaginary root will result in system malfunctions, since that itself gives that 1=-2.

[Narga_Rocks]

Attempting to solve the following equation:

x² + x + 1 = 0

=> x² = - x - 1

=> x = - 1 - 1/x [assuming that x is not equal to 0]

Now, let's substitute this equation for x in the original equation:

x² + (- 1 - 1/x) + 1 = 0

=> x² - 1/x = 0

=> x² = 1/x

=> x³ = 1

=> x = 1

Back to the original equation:

x² + x + 1 = 0

=> 1² + 1 + 1 = 0

=> 3 = 0

=> x = - 1 - 1/x [assuming that x is not equal to 0]

Now, let's substitute this equation for x in the original equation:

x² + (- 1 - 1/x) + 1 = 0

The problem is jumping from here to here while assuming a real solution. x does not actually equal -1 - 1/x if you are assuming a real solution, and so substituting (-1 - 1/x) in place of x when they are not actually equivalent is like subbing any other random thing that doesn't equal x.

I suspect/hope that if you attempt to solve the middle equations assuming a complex answer that you'll get the same x as from the original equation.

Which is

(-1 + isqrt3)/2

(-1 - isqrt3)/2

So really, when he does the substitution step he should say "x = complex number", and then reject any real solutions to the equation. Not that I have any idea how to get (-1 + isqrt3)/2 from x³ = 1.

[peener weener]

Protip: When planning a coup, don't do it in front of the emperor.

*watches as Spike is executed*

Noob.

Emperor of what?

Army?

Weapons?

Empire, even?

None of those.

Emperor in title only.

Emperor in seat.

Emperor in Palace.

Defeated he shall be.

oh hurr

[Narga_Rocks]

Oh

Apparently

((-1 - i(3^0.5))/2)^3 = 1

Who knew?

Seriously, go to http://www.mathsisfun.com/numbers/complex-number-calculator.html and input it.

And x = (-1 - i(3^0.5))/2 really does work for x² + x + 1 = 0.

[Thor Odinson]

Dem hidden complex roots.

[Narga_Rocks]

So x³ = 1 has the roots:

1, ((-1 - i(3^0.5))/2), and ((-1 + i(3^0.5))/2)

learn something new every day.

[Thor Odinson]

Yep.

Makes sense, too. Dem cubics and their three solutions.

[AstraLunaSol]

Vamp's Cloud is the sex.

THAT HAIR I CAN'T STOP LOOKING AT IT ASDFASDFASdAFSDAFSDASFAFDSfADSFADSfADSfa

[El Rey León]

Too much math.

Shakespeare soliloquy tiem

Once more unto the breach, dear friends, once more,

Or close the wall up with our English dead.

In peace there's nothing so becomes a man

As modest stillness and humility,

But when the blast of war blows in our ears,

Then imitate the action of the tiger.

Stiffen the sinews, summon up the blood,

Disguise fair nature with hard-favoured rage.

Then lend the eye a terrible aspect,

Let pry through the portage of the head

Like the brass cannon, let the brow o'erwhelm it

As fearfully as doth a galled rock

O'erhang and jutty his confounded base,

Swilled with the wild and wasteful ocean.

Now set the teeth and stretch the nostril wide,

Hold hard the breath, and bend up every spirit

To his full height. On, on, you noblest English,

Whose blood is fet from fathers of war-proof,

Fathers that like so many Alexanders

Have in these parts from morn till even fought

And sheathed their swords for lack of argument.

Dishonour not your mothers; now attest

That those whom you called fathers did beget you.

Be copy now to men of grosser blood,

And teach them how to war. And you, good yeomen,

Whose limbs were made in England, show us here

The mettle of your pasture; let us swear

That you are worth your breeding- which I doubt not,

For there is none of you so mean and base

That hath not noble lustre in your eyes.

I see you stand like greyhounds in the slips,

Straining upon the start. The game's afoot.

Follow your spirit, and upon this charge

Cry 'God for Harry, England, and Saint George!'

Henry V, bitches.

[Kriemhild]

=> x = - 1 - 1/x [assuming that x is not equal to 0]

Now, let's substitute this equation for x in the original equation:

x² + (- 1 - 1/x) + 1 = 0

The problem is jumping from here to here while assuming a real solution. x does not actually equal -1 - 1/x if you are assuming a real solution, and so substituting (-1 - 1/x) in place of x when they are not actually equivalent is like subbing any other random thing that doesn't equal x.

Yeah, but x = - 1 - 1/x was taken directly from the original equation. There shouldn't be any problem with substituting to the original equation afterwards then. Everything is being worked out from the original equation. We won't know whether they have any real solutions or not unless we try to solve it. This method was a little different from the normal squaring method but it led to x = 1, which isn't an imaginary root.

I suspect/hope that if you attempt to solve the middle equations assuming a complex answer that you'll get the same x as from the original equation.

Which is

(-1 + isqrt3)/2

(-1 - isqrt3)/2

So really, when he does the substitution step he should say "x = complex number", and then reject any real solutions to the equation. Not that I have any idea how to get (-1 + isqrt3)/2 from x³ = 1.

Yes, the roots should be complex but he arrived at "x = 1" nonetheless. You can't just reject that if your method was completely valid. Something must have gone wrong when he was taking that route and I just can't see what it is.

And when I ask for a hint:

Consider this equation:

(1 - x)^1/3 + (x - 3)^1/3 = 1

=> {(1 - x)^1/3 + (x - 3)^1/3}³ = 1³

=> (1 - x) + 3(1 - x)^1/3(x - 3)^1/3{(1 - x)^1/3 + (x - 3)^1/3} + (x - 3) = 1

From the original equation, we know that {(1 - x)^1/3 + (x - 3)^1/3} = 1 so

(1 - x) + 3(1 - x)^1/3(x - 3)^1/3 + (x - 3) = 1

=> 1 - x + x - 3 + 3(1 - x)^1/3(x - 3)^1/3 = 1

=> - 2 + 3(1 - x)^1/3(x - 3)^1/3 = 1

=> 3(1 - x)^1/3(x - 3)^1/3 = 3

=> (1 - x)^1/3(x - 3)^1/3 = 1

=> (1 - x)(x - 3) = 1

=> -x² + 4x - 3 = 1

=> x² - 4x + 4 = 0

=> x = 2

But going back to the original equation:

(1 - 2)^1/3 + (2 - 3)^1/3 = 1

(- 1) + (- 1) = 1

-2 = 1

I get some more complicated stuff that also states that -2 = 1. WHY ARE YOU TORTURING ME SO MUCH?????