## Maybe even more fun than making big numbers

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gmalivuk
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### Maybe even more fun than making big numbers

crazyjimbo wrote:
gmalivuk wrote:pretty please can we be assholes to people who just blindly expect us to jump up and do their work for them?

If they really are saying 'here's my homework, do it', I'm all for being an asshole. My personal preference for assholery would be to give a solution involving large quantities of irrelevant/made up maths, arriving eventually at the the correct solution, and hope they hand it in without really understanding.

I would be most disappointed if others did not continue the discussion with more gibberish maths, a la mornington crescent.

So here's where we answer simple questions with made-up math. I'll start us out with one from the LJ math help community I'm in. Relevant because of how much I hated grading calc assignments where "limits" were computed by calculator.

Find the limit as X approaches 0 for the equation : sin(2X) / X.
On my calculator, the answer is .03491, but the answer should be 2.

When someone decides a question has been suitably answered, ask another one.
Last edited by gmalivuk on Wed Jul 11, 2007 3:58 pm UTC, edited 1 time in total.
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Mouffles
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On a similar note, sin(x)/n = 6. (six)
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Gwydion
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### Re: Maybe even more fun than making big numbers

gmalivuk wrote:Find the limit as X approaches 0 for the equation : sin(2X) / X.
On my calculator, the answer is .03491, but the answer should be 2.

First, set the limit equal to some constant y, which we'll solve for. Next, rearrange so that we get the trig terms by themselves:
lim (sin 2x / x) = y
yx = sin 2x

Now we'll square both sides, giving:
y^2 * x^2 = sin^2 2x

Of course, from trig, we know that we can rewrite the square of a trig function, then put it all onto one side, giving:
y^2 * x^2 = 1 - cos^2 2x
y^2 *x^2 + cos^2 2x - 1 = 0

x = (- cos^2 2 +/- ((cos^2 2)^2 + 4y^2)^0.5) / 2 y^2)

Since there are no imaginary parts to this equation, we can ignore those and take the square root term by term, giving us:
x = (1 / 2 y^2) (-cos^2 2 +/- 2y + cos^2 2)

The cosine terms cancel out *finally*, so we can start to evaluate this.
x = +/- (2y / 2y^2)
x = +/- (1 / y)

Rearrange (remember, we're solving for y), giving us:
y = +/- (1 / x)

But since we're taking the limit as x goes to 0, this goes to positive and negative infinity at the same time. That's just plain impossible, so the limit must not exist.

How did you get 2 again?

[Note: I used to TA for college algebra, and I promise that every mistake made above was handed in on homework no less than four times a semester.]

Twasbrillig
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Mouffles wrote:On a similar note, sin(x)/n = 6. (six)

Very original. :3
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### Re: Maybe even more fun than making big numbers

Gwydion wrote:How did you get 2 again?

It's easy to get 2. By the linearity of sine, lim x->0 sin(2X)/X = lim x->0 2sin(X)/X = 2 x lim x->0 sin(X)/X = 2 x 1 = 2.

Heh, Now I think about it, thats not *too* far from the truth.

parallax
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sin(2X)/X = (sin2)X/X, by associate law.
(sin2)X/X = (2sin)X/X, by commutative law.
(2sin)X/X = (2sin)1, by multiplicative inverse law.
(2sin)1 = 2(sin1), by associative law.
2(sin1) = 2(1), 1 is positive, so it's sign is +1.
2(1) = 2. Multiplication.
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Gwydion
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That's way too close to being the right answer for my liking. You didn't add any roots to an equation, or try to invoke proofs of 0 = 1. Disappointing.

I think guess and check should suffice, then:

sin(4pi)/2pi = 0
sin(3pi)/1.5pi = 0
sin(2pi)/pi = 0
sin(pi)/0.5pi = 0

Looks to me like this trends toward zero. It even does from the other side, if you don't believe me.

parallax
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Canceling the X's, you get sin(2X)/X = sin 2, which is obviously "Making for yourself an idol, whether in the form of anything that is in heaven above, or that is on the earth beneath, or that is in the water under the earth."
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Aldarion
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As I see it, the question has been thoroughly answered, so I'll ask a new one:

Prove, in the most convincing way, that 169 is prime.
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Token
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Assume 169 is not prime. Then there exist positive non-one integers t and h such that 169 = th. Let r and e equal one. Then 169 = three, which is a contradiction. Therefore 169 is prime.

Gwydion
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Isn't this just like the divisible-by-three test?

1+6+9 = 16
1+6 = 7

Since 7 is prime, 169 must be prime.

Check with 170:
1+7+0 = 8, which is composite. The theory checks out.

gmalivuk
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If the Riemann Hypothesis is true, all nontrivial zeroes of the zeta function have real part equal to 1/2. As von Koch proved in 1901, this is clearly equivalent to saying there is a constant C such that for all sufficiently large x.

From this, it becomes a fairly simple exercise (#7 in the problem set) to show that 169 is prime.

Unfortunately, though, we cannot depend on the truth of Riemann. If the Hypothesis is false, there is no such constant C above. However, it is also then true that does *not* hold for all ε > 0. We can therefore choose some ε0 for which the relation does not hold. Substituting x = 169 in this equation, it is again a simple exercise (problem #15) to show how this implies 169's primality.
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parallax
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Are we allowed to assume the prime numbers are closed under multiplication?
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Yakk
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By fermat's little theorem, for all primes p and integers a, a^(p-1) == 1 mod p.

Let a = 12. Then,
12^168 mod 169
==
(12^13)^12

Now,
12^1 == 12
12^2 == 144
12^4 == 118
12^8 == 66
so 12^13 == 12^8 * 12^4 * 12^1 == 66*118*12 == 93456.

As 169*552 + 168 = 93456, we have:

(12^13)^12 mod 169
==
(168)^12
==
(-1)^12
==
((-1)^3)^4
==
(-1)^4
==
1^2
==
1

Examining the list of charmicheal numbers, 169 does not appear. As such, 169 is prime.

Freddino18
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### Re: Maybe even more fun than making big numbers

prove that 2 + 2 = 5

Last bumped by Freddino18 on Thu Jan 17, 2019 5:01 pm UTC.

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