Where the hell did you hear this? Maybe someone misinterpreted the definition of a derivative perhaps, or the limit. Zero is the identity element of addition. a + b = a, so we call b zero.. nothing.. nada. But are you saying 0/infinity = 1 or x/0^infinity = 0? Maybe... Limit(x->0) x/(1/x) which would look like 0/oo but it is this: Limit(x->0) x^2 = 0, not one.
Nice semi-proofs, przyk. Of course, in the first one, fg'' does not equal 4(x^9)e^(x^2), so you're not actually comparing two terms of f'g'. fg''= 2(x^7)e(x^2) + 4(x^9)e(x^2), which acounts for the difference between the (f'g')s The second one is even more simple. Still, they're quite cute, though, and I hadn't seen the first one. I'll have to try out the first one on a couple of friends...
A more suble one: Let u = 1 + <sup>1</sup>/<sub>3</sub> + <sup>1</sup>/<sub>5</sub> + <sup>1</sup>/<sub>7</sub> + <sup>1</sup>/<sub>9</sub> ... and v = <sup>1</sup>/<sub>2</sub> + <sup>1</sup>/<sub>4</sub> + <sup>1</sup>/<sub>6</sub> + <sup>1</sup>/<sub>8</sub> + <sup>1</sup>/<sub>10</sub> ... Now, 2v = 1 + <sup>1</sup>/<sub>2</sub> + <sup>1</sup>/<sub>3</sub> + <sup>1</sup>/<sub>4</sub> ... = u + v, therefore u = v or u - v = 0 If we calculate u - v term by term, we get: (1 - <sup>1</sup>/<sub>2</sub>) + (<sup>1</sup>/<sub>3</sub> - <sup>1</sup>/<sub>4</sub>) + (<sup>1</sup>/<sub>5</sub> - <sup>1</sup>/<sub>6</sub>) ... Since all the terms in this summation are greater than zero (1 > <sup>1</sup>/<sub>2</sub>, <sup>1</sup>/<sub>3</sub> > <sup>1</sup>/<sub>4</sub>, and so on) we can safely conclude that u - v > 0. Substituting u - v gets us 0 > 0. The inescapable conclusion is that zero is strictly greater than itself (ie. equality is excluded).
Yea, strange things seem to happen with infinite series. It's like.. in your "proof" you assume that (1 + 1/3 + 1/5 + 1/7 + ....) + (1/2 + 1/4 + 1/6 + ...) makes sense if you move the terms in v an infinite number of "places" to the left so we can group them such that u+v = 1 + 1/2 + 1/3 + ... Silly Billy
And a dumb one (this one's my own idea): It's quite well known that, when multiplying odd and even numbers: even * even = even even * odd = even odd * even = even odd * odd = odd Let p and (1-p) be the proportion of odd and even numbers in the set of integers. The above table yields: Proportion of even numbers: (1-p)<sup>2</sup> + 2p(1-p) Proportion of odd numbers: p<sup>2</sup> The simultaneous equations p = p<sup>2</sup> and (1-p) = (1-p)<sup>2</sup> + 2p(1-p) allow only two solutions, namely: p = 0 or p = 1. This debunks the myth that there are as many even as odd integers.
What? I can put them where I want. Addition is commutative, remember? In fact, I'm pretty sure that the sum of two series is defined as the series of the sums of the corresponding terms.
This old concept again. I'm no math whiz. But from reading through these threads in the past, it's pretty much a no-brainer. 0.9999.... is 1. It's in the very definition. The decimal 0.999.... is the real number that is the limit of the sequence beginning 0.9, 0.99, 0.999, ... That real number is 1.
I remember seeing an example somewhere where it doesn't work... it's in a book I own. If I find it, I will let you know. But a different mistake is how you arrived at u-v = 0.. which isn't true.
No doubt what Absane is recalling is the fact that one can only perform an infinite rearrangement of the terms of an infinite series if the series is absolutely convergent. A famous example of the dangers of rearrangement is the alternating harmonic series Σ (-1)^n 1/n . The series converges, but it doesn't converge absolutely. It isn't hard to prove that you can make the series sum to anything you want by rearranging terms! Notice that przyk's two series don't converge at all! P.S. My favorite infinite series: Σ n = -1/12. Figure that out!
1.000000 ^ 1000000=1 .999999^1000000=.367879 1/(.999999^1000000) or 1.0000001^1000000=2.71828 and change. 2.71828=e, the base of the natural logarithms.
Lol. Thank you sir.. it's what I was trying to get two but it's been 3 years since I took Calc 2. Sure, the book is within reach but I would rather just pretend I know Please Register or Log in to view the hidden image!
u and v do not converge, and when you do u-v, you are really saying Infinity - Infinity = 0.. which isn't true. However, I guess that is the same reasoning all along. Dang.
Let me make something up Please Register or Log in to view the hidden image! 1/12*{Floor[cos(pi*1] + Floor[cos(pi*1/2]+Floor[cos(pi*1/3)] + ...}
The fact that u and v do not converge in itself isn't the problem; it just means that you have to be careful by what you mean when you compare non convergent series. Hint: If you define addition and subtraction of two series like this: ∑<sub>i</sub> a<sub>i</sub> + ∑<sub>i</sub> b<sub>i</sub> = ∑<sub>i</sub> (a<sub>i</sub> + b<sub>i</sub>) and ∑<sub>i</sub> a<sub>i</sub> - ∑<sub>i</sub> b<sub>i</sub> = ∑<sub>i</sub> (a<sub>i</sub> - b<sub>i</sub>), what's the corresponding definition of equality between two infinite series going to be?
