An argument of infinite proportions (pi)
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Please by all means message this joke to me! Unless its the one about the Special Olympics of course lol, because that is all too familiar.
Masta Squidge
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My thinking is that pi, must in fact, end .
(for it must provide a shape).
Whilst 1/3 does not have to provide a shape.
it is not his work.
can dream.JuanV.Soler
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JuanV.Soler wrote:
My thinking is that pi, must in fact, end .
(for it must provide a shape).
Whilst 1/3 does not have to provide a shape.
it is not his work.
can dream.Smile
ooh circular reasoning (pun intended)
That makes sense, however there is nothing to say that a ratio cannot work properly without an end, only infinitely increasing levels of accuracy (which on its own is a paradox, because you cant do better than perfect right?)
I would be interested in running that by someone in the profession of calculating stupid numbers to places beyond any usable level of accuracy. Something about that statement, while it makes sense right now, doesn't exactly sit right in the back of my mind and I cant place it.
Anyways, I'm off to bed so I can survive work tonight, please by all means continue this because I love reading these sorts of things.
Masta Squidge
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Quote:
That makes sense, however there is nothing to say that a ratio cannot work properly without an end, only infinitely increasing levels of accuracy (which on its own is a paradox, because you cant do better than perfect right?)Sorry, I dont get you.
JuanV.Soler
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So who was the genius that coined the phrase "easy as pi"?
solo
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A number isn't just something made up it's the answer to an equation. Under that definition if you take that answer to equations can sometimes be quantified as between a real & imaginary number and infinity then numbers can go on forever. Numbers can be limitless.
You should tell him 2 isn't really 2 and see what he says.
Pi is limitless, Pie is not... especially around guys from the south.solo
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Okay, maths is very much my weakness, but I'm confused by your stance. In the other forum you wrote:
"I say they never end as pi is an irrational (which by definition never ends) and that 1/3 also never ends as it is a non terminating repeating decimal."
But in your post here you state:
"Now, anyone who paid attention in math class in middle school knows that he is wrong, that pi is called an irrational number for a reason, and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."
For the record I agree that neither number ends- especially 1/3 as a decimal. Your "opponent" ain't worth the argument if he seriously reasons that 1/3 expressed as a decimal logically ends in a 3!
Jackson
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What are you guys smoking? E Grin Pi r not Sq...Pi art round! Thumbs up
3eighty
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Masta Squidge wrote:
someone in the profession of calculating stupid numbers...Recently it was reported that some university professor here in Canada completed a massive seven or eight year long calculation using a supercomputer. He was able to absolutely prove that a game of checkers will always end in a draw provided neither player ever makes a mistake or cheats. It is only the mistakes and cheats that swing the outcome away from a draw.
Its great to see my tax dollars at work.
I can't understand why anyone would think 1/3 would ever end. The repeating character of the ratio seems easy enough to comprehend as unending. I hope that university professor doesn't decide to tackle the 1/3 number next.
Ross Macintosh
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Jackson Barkess wrote:
...and that 1/3 does in fact end, thats why its called a repeating, terminating decimal."I believe that is what we call a typo Embarassed I believe it should read "does in fact NOT end, thats why its called a repeating, NON-terminating decimal.
Yeah, I slipped, it was past my bedtime.
Masta Squidge
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Masta Squidge wrote:
Please by all means message this joke to me! Unless its the one about the Special Olympics of course lol, because that is all too familiar.Yep, that's the one.
SchreiberBike
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Not that it will help, but here is a mathematical proof that pi is irrational:
Quote:
Niven's proofThe proof uses the characterization of π as the the smallest positive zero of the sine function. Like all proofs of irrationality, the argument proceeds by reductio ad absurdum.
Preparation: Suppose that π is rational, i.e. π = a / b for some integers a and b ≠ 0, which may be taken without loss of generality to be positive. Given any positive integer n, we define the polynomial function
f(x) = {x^n(a - bx)^n \over n!},\quad x\in\mathbb{R},
and denote by
F(x) = f(x) + \cdots + (-1)^j f^{(2j)}(x) + \cdots + (-1)^n f^{(2n)}(x). ,
the alternating sum of f and its first n even derivatives.
Claim 1: F(0) = F(π)
Proof: Since
f(x)=b^n{x^n(\pi - x)^n \over n!}=f(\pi-x),\quad x\in\mathbb{R},
the chain rule and mathematical induction imply
f^{(j)}(x) = (-1)^j f^{(j)}(\pi - x),\quad x\in\mathbb{R},
for all the derivatives, in particular
f^{(2j)}(0)=f^{(2j)}(\pi),
for j = 1, 2, ...,n and Claim 1 follows from the definition of F.
Claim 2: F(0) is an integer.
Proof: Using the binomial formula to expand (a – bx)n and the index transformation j = k + n, we get the representation
f(x)={1\over n!}\sum_{j=n}^{2n}{n \choose j-n}a^{2n-j}(-b)^{j-n}x^{j}.,
Since the coefficients of x0, x1, ..., xn − 1 are zero and the degree of the polynomial f is at most 2n, we have f (j)(0) = 0 for j < n and j > 2n. Furthermore,
f^{(j)}(0)={j!\over n!}{n \choose j-n}a^{2n-j}(-b)^{j-n}\quad\mbox{for } n\le j\le 2n.
Since j ≥ n, the fraction of the two factorials is an integer. The same holds for the binomial coefficient, as can be seen from its combinatorical interpretation or Pascal's triangle. Hence f and every derivative of f at 0 is an integer and so is F(0).
Claim 3:
\frac12 \int_0^\pi f(x)\sin(x),dx=F(0)
Proof: Since f (2n + 2) is the zero polynomial, we have
F'' + F = f.,
The derivatives of the sine and cosine function are given by (sin x)' = cos x and (cos x)' = −sin x, hence the product rule implies
(F'\cdot\sin - F\cdot\cos)' = f\cdot\sin.
By the fundamental theorem of calculus
\frac12 \int_0^\pi f(x)\sin(x),dx= \frac12 \bigl(F'(x)\sin x - F(x)\cos x\bigr)\Big|_{x=0}^{x=\pi}.
Since sin 0 = sin π = 0 and cos 0 = –cos π = 1 (here we use the abovementioned characterization of π as a zero of the sine function), Claim 3 follows from Claim 1.
Conclusion: Since f(x) > 0 and sin x > 0 for 0 < x < π (because π is the smallest positive zero of the sine function), Claims 1 and 3 show that F(0) is a positive integer. Since
x(\pi -x) = \Bigl(\frac\pi2\Bigr)^2-\Bigl(x-\frac\pi2\Bigr)^2\le\Bigl(\frac\pi2\Bigr)^2,\quad x\in\mathbb{R},
and 0 ≤ sin x ≤ 1 for 0 ≤ x ≤ π, we have
\frac12 \int_0^\pi f(x)\sin(x),dx\le \frac{b^n}{n!}\Bigl(\frac\pi2\Bigr)^{2n+1},
which is smaller than 1 for large n, hence F(0) < 1 by Claim 3 for these n. This is impossible for the positive integer F(0).
Not that I understand it, but there it is.
SchreiberBike
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Aha...
Well, thanks - at least for that very last line (now I can hope I'm not alone in the universe who does not understand itGaieus
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