19 Nov 14
Originally posted by Great King RatMath has nothing to do with arrogance. Math is an exact science. Every step is provable, if not it is just a hypothesis. So when you say that 1/2 + 1/4 + 1/8 + ... is not exact 1, and it is your opinion and you cannot prove it nor understand the stringent prove, and then ask us to agree with you - that's arrogance.
Yeah... um... apparently you feel the need to be arrogant and condescending to me.
19 Nov 14
Originally posted by Great King RatI've been thinking about how to explain this. The series 1 + x + x² + ··· is the limit of a sequence of polynomials. The nth polynomial is P(x; N) = 1 + x + ··· + x^N = (1 - x^(N+1))/(1 - x), which can be proved in a number of ways. For example, suppose the equation is true for the Nth case then for the case N + 1 we would have:
So earlier you wrote:
"instead, we define the sum to be the number that the sequence converges to."
So, in other words: "we define the sum to be the number that the sequence gets closer and closer to."
I agree with this.
I still don't see why we therefore should say "at infinity the number the sequence [b]is is 1", when w ...[text shortened]... nity the number the sequence gets infinity close to is 1".
Sorry, I just don't get it.[/b]
P(x; N + 1) = 1 + x + x² + ··· + x^(N + 1)
P(x; N + 1) = P(x; N) + x^(N + 1)
P(x; N + 1) = (1 - x^(N + 1))/(1 - x) + x^(N + 1)
And so:
P(x; N + 1) = (1 - x^(N + 1) + [x^(N + 1) - x^(N + 2)])/(1 - x)
P(x; N + 1) = (1 - x^(N + 2))/(1 - x)
which is what we would have obtained if we had just substituted N + 1 into the formula. In other words if the formula is true for one polynomial in the sequence it is true for the next. In the case P(x; 0) we have:
P(x; 0) = (1 - x^(0 + 1))/(1 - x) = 1
and, looking at the polynomial:
P(x; 0) = [1 + x + x² + ··· + x^(N + 1)]|(N = 0)
P(x; 0) = 1.
So the formula works for N = 0 and by iteration all subsequent cases. We therefore have verified the formula:
P(x; N) = 1/(1 - x) - x^(N + 1)/(1 - x)
So, how about the infinite series. If |x| < 1 then as N increases the second term becomes smaller and smaller. So we might guess that only the first term survives. Can we justify that statement. First I need to introduce a mathematical fact about the real numbers - they are complete meaning there are none missing. This is not true of the integers as one can ask the question what is between 2 and 3 and not be able to find an integer that answers that question. It is also not true of the rationals as one can find numbers like sqrt(2) which are 'missing'. For our series to not equal 1/(1-x) there should be a number Å‹ such that P(x;N) < 1/(1 - x) - Å‹ < 1/(1 - x) for all N. Clearly if N = 3 (say) we can find such a number. For clarity I'll specialise to the case x = ½. Then:
P(½; 3) = 1 + ½ + ¼ + â…› = 2 - â…›
So Å‹ = 1/16 will fit neatly into the gap between 2 and 2 - â…›. However if we add a couple more terms then P(½; 5) = 2 - 1/32 and Å‹ is too big. In fact for any Å‹ > 0 we care to choose we can always find an N such that P(½; N) > 2 - Å‹. We can work out what N is for any Å‹:
P(½; N) = 2 - ½^(N + 1) > 2 - Å‹
=> ½^(N + 1) < Å‹
=> N + 1 > log(Å‹)/(log(½ )
=> N > log(Å‹)/log(½ ) - 1
Note that log(Å‹) and log(½ ) are both negative which is why I had to swap around the inequality. This means that there is no number, no matter how close we make it to 2, such that we can't find a polynomial closer to 2. Since the series is larger than any of the partial sums, and the real numbers are complete the series cannot be smaller than 2.
19 Nov 14
1 edit
Originally posted by adam warlockWith stereographic projection, when mapping the complex plane to the Riemann sphere, there is the problem of what to do with the north pole at (0,0,1). We could just map (0,0,1) to "kumquat" and be done with it, but as luck would have it "infinity" was chosen, depicted by an 8 knocked onto its backside. Big deal. And where's infinity on the resultant extended complex plane? Unlike infinity on the extended real line, there is no one particular direction where it lies. It's "in every direction." So infinity is a device that's handy for mathematicians and physicists, but I can see how the casual observer on the street might think it looks like a hand-waving cheat.
They use it in very specific formal contexts:
The extended real number line
Non-standard analysis
In complex analysis you have the concept of riemann surfaces that really only work well with infinity
etc.
In an informal way mathematicians say "at infinity" a lot but what they usually mean to say is that we are considering the behavior of mathematical entities at larger and larger values...
19 Nov 14
The post that was quoted here has been removedWe could use up two Eternities in learning all that is to be learned about our own world and the thousands of nations that have arisen and flourished and vanished from it. Mathematics alone would occupy me eight million years. - Mark Twain
I think Mark Twain was being overly optimistic.
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20 Nov 14
1 edit
The post that was quoted here has been removedI thought ( and may be wrong ) that the imaginary numbers are on the imaginary number line ( or plane ) which is an entirely different set outside of the Reals ( unlike integer,rational,irrational which are subsets of the Reals)
In fact shouldn't the Reals just be a subset of the Imaginay numbers, since the Real number line is just an axis in the Imaginary plane?
I'm having trouble following your argument.
20 Nov 14
2 edits
The post that was quoted here has been removedI should probably have left the proof by iteration out, but having written it didn't really want to delete it. By completeness I meant in the sense that any non-empty subset of the reals has a least upper bound which is a real, the example they give on the Wikipedia page is {x in Q | x² < 2} where Q is the set of rationals has least upper bound sqrt(2) but sqrt(2) is irrational so the set Q is incomplete. The completeness argument is necessary as GKR may think that in some way we have a number missing, the discussion about the extended real numbers introduces the notion of points at infinity, and one can also add infinitesimals - smaller than any real number but bigger than zero. With that kind of construction one could introduce an argument that 0.999··· is infinitesimally smaller than 1.0, which it is not so it was a matter of closing a loophole.
In the meantime I've had another idea as to how to explain it.
20 Nov 14
2 edits
Alternative explanation:
If a and b are distinct then there is another number (a + b)/2 between them, what is more (a + b)/2 is distinct from both a and b. If a = (a + b)/2 or b = (a + b)/2 then a = b.
Lets try working out (a + b)/2 for a = 1.0 and b = 0.999···
(a + b)/2 = (1.0 + 0.999··· )/2
(a + b)/2 = 1.999···/2
Doing the long division we get 0.999··· or in other words the average of the two numbers is equal to one of the numbers and so 0.999··· and 1.0 are equal.
20 Nov 14
Originally posted by DeepThoughtPlease tell us more about this 'completeness'. The completeness I learned in school had the integers being a complete set.
First I need to introduce a mathematical fact about the real numbers - they are complete meaning there are none missing. This is not true of the integers as one can ask the question what is between 2 and 3 and not be able to find an integer that answers that question.