John

Once when I was a teenager, I cut my finger such that it started bleeding pretty heavily (more than I would have expected for what was a fairly small, if deep, cut). I had no clue about first aid back then, and my attempts to stop the bleeding just seemed to make it worse. It was very early in the morning, so I didn't want to wake my parents. After a few minutes, though, I got to the point of feeling woozy, disoriented and tired. I panicked a bit and tried calling out for help, but I found myself unable to really put any force behind it. So I just sort of stood there, leaning against the counter, bleeding into the sink, too exhausted to do much but also not terribly concerned. Fortunately my dad woke up and came out not too long afterwards, and he was quickly able to stop the bleeding.

What I experienced certainly falls into the range of signs of hypovolemic shock (which Dad, a former paramedic, mentioned at the time), though other factors besides blood loss may have played a part (especially since I'm not entirely convinced I lost that much blood). In any case, that was my experience. I imagine that if Dad hadn't come out when he did, I would have fainted before long.

Alright, having looked into this a bit more (namely on Wikipedia here and here), here's a derivation.

First, we have that [math]\Gamma(1 - x) = -x \Gamma(-x)[/math].

Using Weierstrass factorization, we have

[math]\sin{\pi x} = \pi x \prod_{n=1}^{\infty} \left(1 - \frac{x^{2}}{n^{2}}\right)[/math]

and

[math]\frac{1}{\Gamma(x)} = xe^{\gamma x} \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right)e^{-\frac{x}{n}}[/math].

This is enough to derive the formula, as follows:

[math]\begin{array}{rcl}\frac{1}{\Gamma(x) \Gamma(1 - x)} & = & \frac{1}{-x\Gamma(x) \Gamma(-x)} \\ & = & \left(\frac{1}{-x}\right) \left(xe^{\gamma x} \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right)e^{-\frac{x}{n}}\right) \left(-xe^{-\gamma x} \prod_{n=1}^{\infty} \left(1 - \frac{x}{n}\right)e^{\frac{x}{n}}\right) \\ & = & \left(\frac{1}{-x}\right) \left(xe^{\gamma x}\right) \left(-xe^{-\gamma x}\right) \prod_{n=1}^{\infty} \left(1 + \frac{x}{n}\right)e^{-\frac{x}{n}}\left(1 - \frac{x}{n}\right)e^{\frac{x}{n}} \\ & = & x\prod_{n=1}^{\infty}\left(1 - \frac{x^{2}}{n^{2}}\right) \\ & = & \frac{\sin{\pi x}}{\pi}\end{array}[/math].

Since [math]\frac{1}{\Gamma(x) \Gamma(1 - x)} = \frac{\sin{\pi x}}{\pi}[/math], we have [math]\Gamma(x) \Gamma(1 - x) = \frac{\pi}{\sin{\pi x}}[/math].

I'm not sure what you mean by that, since in both the ProofWiki article and the PDF, the derivations are explained step-by-step. The author of either one could have made mistakes somewhere along the line, which I'm not qualified to judge since I'm not familiar with the requisite concepts and I'm not particularly inclined to spend that much time on learning them right now, but they certainly don't lack explanation.

This is Euler's reflection formula, and a derivation can be found at ProofWiki: https://proofwiki.org/wiki/Euler's_Reflection_Formula

A more detailed derivation starts near the bottom of page 5 here: http://homepage.tudelft.nl/11r49/documents/wi4006/gammabeta.pdf

Both use concepts from complex analysis, of which I know very little. So perhaps someone more knowledgable can provide some insight. But anyway, there are two starting points if you'd like to explore further.

Well, [math]|x| = \sqrt{x^{2}}[/math] in any case, so changing from one to the other won't affect much.

As for the rest, at a quick glance, I don't think modular arithmetic makes things terribly messy. For differentiation and integration, taking the derivative (or integral) of the starting function and then replacing instances of x with x mod n would probably work, and I'm not sure where the problem would lie with identities.

I haven't looked into it terribly much, but one way we can generate something periodic is by using modular arithmetic, e.g. check out

[math]y = \Gamma (x \textnormal{ mod } 2)[/math]: http://www.wolframalpha.com/input/?i=y+%3D+gamma(x+mod+2)

and

[math]y = \Gamma \left(-(x \textnormal{ mod } 2)^{2}\right)[/math]: http://www.wolframalpha.com/input/?i=y+%3D+gamma(-(x+mod+2)^2)

Edit: I'm assuming that instead of "rotational symmetry," you meant "translational symmetry," since the latter is closer to what periodicity implies, but if you do want rotational symmetry, more elaborate constructions may be required. For example, going back to the gamma function, we can do something like [math]y = \frac{x}{|x|} \Gamma(|x|)[/math]: http://www.wolframalpha.com/input/?i=y+%3D+%28x%2F%7Cx%7C%29gamma%28%7Cx%7C%29

But, if we visualise this graph by extrapolating backwards from a universe with accelerating expansion, the line forms an asymptote with the x axis pushing the origin back to infinity.

 

So how can there have been a singularity? Or how can the universe be finite in age?

Others with greater knowledge of cosmology have provided good information, but I'd like to chime in here to say this reasoning is a bit flawed, in a sense seemingly similar to that employed by Zeno in his paradoxes.

As a counterexample, consider the graph of y = x², and imagine we're living on the curve at the point (100, 10000). Observations made at various points on the curve will show us that y is growing, and the growth is accelerating, but looking back to x = 0 (assuming we can), we'll see that y = 0 as well, and furthermore, this point is a finite distance (100 units) away along the x axis. That is to say, accelerating expansion does not imply that the size of the universe was never zero (though other physical considerations may).

Of course, I may have misunderstood your post. In that case, disregard.

Depending on what exactly your instructor expects, this may be a valid answer.

However, a couple of things:

1. Formally, using the definition, the statement is true, but is O(x²) the tightest bound?

2. Keep in mind that c and k aren't restricted to being natural numbers. I mention this only because it seems like you're looking for the smallest values.

The video's good for a bitter laugh, but it doesn't present anything new or compelling. It rehashes the same tired misinformed arguments intelligent design proponents have been spouting for years.

I skipped to the 10-minute mark, as you indicated, and almost immediately heard the guy claim that evolution is supposed to explain the origin of life. This is false. The origin of life from non-living material is called abiogenesis, and while we have a few ideas of how it might have happened, we don't know for sure. Meanwhile, evolution describes how diversity and speciation arise among living organisms, and modern evolutionary theory is about as close to absolute truth as science gets.

I actually listened to a few more minutes of the video, as he spouted off about how there are no transitional fossils, and how our inability to directly observe evolution means we can't accurately estimate the timescales on which it operates, and how modern dating methods are wrong for various reasons (and he seems to believe absolute and relative dating are one and the same), etc. It quickly became apparent that the creator of the video has several fundamental misunderstandings of evolutionary theory. Moreover, he doesn't seem like the type to be swayed by reasoned argument supported by evidence.

I wouldn't bother engaging with this person, except perhaps to recommend that he take an introductory biology course or two. But then, I doubt that would convince him either.

I think you may be misreading his post. The relevant thread of conversation is as follows:

But that's what I mean, would time dilation actually create a delay in the nerve signals?

followed by

In principle, yes.

which answers your question, but then he makes the note that

In practice your arm would be torn off before you noticed.

The function y = (floor(x))^2 isn't continuous. See the graph at WolframAlpha: http://www.wolframalpha.com/input/?i=y+%3D+floor%28x%29%5E2

As Bignose mentioned, the derivative of the floor function is 0 for all non-integer x and undefined for all integer x. Thus, using the chain rule, we see that the derivative of floor(x)^2 = 2floor(x) * 0 = 0 for non-integer x, and of course it's undefined for integer x.

Also, as a minor nitpick, y = x^2 isn't monotonic.

Edit: My second paragraph is worded a bit poorly, as it seems to imply that non-differentiability implies discontinuity, which is certainly false (in fact, most continuous functions are nowhere differentiable). But since the derivative of floor(x)^2 was mentioned earlier, I'll leave it in anyway, for its general point.

But Strange answered that question in post 16 (see his first three words), and you just said that was "a copout answer".

Of course, maybe you just meant that the "in practice" part was a copout, though I don't see why you'd say that; but in any case, you didn't indicate that his answer was what you were looking for, though now it seems to have been just that. That's probably why other members assumed you were looking for something different, which seems to have led to some confusion.

Keep in mind that my response here is in the context of the real numbers with addition and multiplication defined in the usual way.

The answer is yes!

infinity*0= infinity (1-1)=infinity-infinity, which equals any number. because

infinity-infinity-3 is absorbed in infinity like a blackhole. and still equals infinity-infinity, likewise infinity-infinity-5 equals the same thing.

The answer is yes, but not for the reason you claim.

Infinity is not a number, and thus arithmetic statements containing infinity, like [math]\infty - \infty[/math], aren't valid.

Rather, something like [math]0 \times \infty[/math] comes up in the context of some limiting process. The "indeterminate" aspect can be thought of as arising because we can take different "paths" towards [math]0 \times \infty[/math] depending on the limit in question, and arrive at different results.

Consider, for example, the following four limits, which all approach [math]0 \times \infty[/math] in the limit:

1. [math]\lim_{x \to \infty}0 \times x = 0[/math]

2. [math]\lim_{x \to \infty}\frac{1}{x} \times x = 1[/math]

3. [math]\lim_{x \to \infty}\frac{1}{2x} \times x = \frac{1}{2}[/math]

4. [math]\lim_{x \to \infty}\frac{1}{x} \times x^2 = \infty[/math]

"99.9% of infinity" isn't really valid, but if it were, then yes.

Just moved to Australia several months ago, still getting used to people using "40 degrees" to describe a hot day, or seeing speed limits over 100 (kph), or adults talking about weighing "70" without being severely underweight, etc. Overall, though, it's not too big an adjustment, though I guess the fact that I've been a science nerd for ages has helped.

Anecdote: Not too long ago I went to the gas station ("servo") down the street to get a few things. It was late, so the doors were locked and I had to request the items through the service window outside. I asked the girl for a quart of milk. (Admittedly mild) hilarity ensued.

As mentioned previously in the thread, consider that light travels at around 300,000,000 m/s while nerve impulses from the hand travel closer to 60 m/s. Thus, even without taking time dilation and such into account, the light from a person's hand touching an object always reaches the eyes "long" before the nerve signal from the hand reaches the brain. At normal human scales, the difference isn't great enough to be really noticeable, especially since there are probably differences in how long the brain takes to register sight versus touch.

The effect would, of course, be magnified in the case of a person 100 billion light-years tall. Assume each arm would then be something like 50 billion light-years long, and we get the light from the tip of the hand taking about 50 billion light-years to reach the eye, with the nerve impulse taking closer to 250 quadrillion years to reach the brain.

While your OP does ask for in-thread explanations, you might enjoy these Wikipedia articles:

http://en.wikipedia.org/wiki/Introduction_to_special_relativity

http://en.wikipedia.org/wiki/Introduction_to_general_relativity

They provide decent overviews while remaining fairly accessible.

Dunno, the wording of the first line of the OP seems to indicate that, whether he means rationals or reals, he intends for both intervals to be subsets of the same set. But perhaps he'll respond and say otherwise.

 

Yes so long as you do not try to make injections from the reals to the rationals.

Certainly.

That proves that there are at least as many numbers between 0 and 1 as there are between 1 and infinity.

There are , I think, actually rather more

http://en.wikipedia.org/wiki/Aleph_number

No, the cardinality of (0, 1) is the same as the cardinality of (1, infinity). The fact that there exists a bijection between the two intervals is proof.

 

I did wonder at the OP's use of fractions since it implies his underlying set is the set of rational numbers, not the reals.

 

Which is intended needs to be made clear before constructing a proof.

The same reasoning applies with the rationals.

Edit: To add more detail, the function in question is f : (0, 1) -> (1, infinity) defined by f(x) = 1/x. The construction of the rationals and the reals ensures that in either set, every non-zero element has a multiplicative inverse. Furthermore, since every x will be less than 1 and positive, every y = 1/x will be greater than 1 and positive. Thus, our function is valid on (0, 1) in either the rationals or the reals, i.e. every x in (0, 1) will be mapped to some y = 1/x in (1, infinity).

Consider two distinct elements y₁ = 1/x₁ and y₂ = 1/x₂ such that y₁ = y₂. This means 1/x₁ = 1/x₂, therefore x₁ = x₂. Thus the function is injective.

Now take any y in (1, infinity). If y = 1/x, then we have x = 1/y. Since every non-zero element of the reals (or rationals) has a multiplicative inverse, this x certainly exists. Furthermore, since y > 1 and the positive reals (or rationals) are closed under multiplication, 1/y must be less than 1 and positive, i.e. 1/y = x must be in (0, 1). Thus the function is surjective.

Since the function is injective and surjective, it is bijective by definition. And since the existence of a bijection between two sets implies the two sets are equinumerous, we have that (0, 1) and (1, infinity) are equinumerous.

The reasoning you just used is essentially the proof. By pairing each real number in [math](1, \infty)[/math] with exactly one number in [math](0,1)[/math] (taking into account, of course, the equivalence between certain numbers in the sets, e.g. 6/4 = 3/2), you've constructed a bijection (specifically, the mapping f(x) = 1/x) between the two intervals, which in turn means they have the same cardinality, i.e. the same number of elements.

In fact, when dealing with the real numbers (or even just the rationals), there are as many elements in any non-empty, non-degenerate interval as there are in any other non-empty, non-degenerate interval.

Edit: To be clear, pairing each real number in [math](1, \infty)[/math] with exactly one number in [math](0,1)[/math] forms an injection, but it's easy to show the mapping in question is also surjective, thus it is a bijection.

Yes, the formula works fine, but again, knowing the fraction is an integer involves knowing we can reduce it, which requires that we already know whether x is prime.

This is certainly true, but knowing whether the fraction is in lowest terms would involve knowing whether x is prime in the first place. Also, x can be 2, as our fraction then becomes 2/1! = 2/1, which is in lowest terms; though of course, we already know 2 is prime.

So, I've proposed another conjecture that all hailstone sequences start with an odd number that contains a factor of 3. This is because given any odd number, we can reverse the algorithm to determine all of the odd numbers that will reduce to the given odd number. Once you reach an odd number that contains a factor of three, we can go no further in reversing the algorithm. Thus, suggesting my conjecture is true.

It is certainly true, though usually we allow the term "hailstone sequence" to apply with any starting number, e.g. {6, 3, 10, 5, 16, 8, 4, 2, 1} is a hailstone sequence.

Anyway, consider that any odd natural number n is congruent to 0, 1 or 2 (mod 3). In order to show that your conjecture holds, we simply must show for each case whether (n * 2^x) - 1 = 0 (mod 3) (where x is, of course, a natural number).

Case 0: Since n = 0 (mod 3), we have 2n = 0 (mod 3), 2(2n) = 0 (mod 3), etc. That is, subtracting 1 always yields 2 (mod 3), i.e. for all x, 2^xn - 1 = 2 (mod 3) and as such is not divisible by 3.

Case 1: Since n = 1 (mod 3), we have 2n = 2 (mod 3), 2(2n) = 1 (mod 3), etc. That is, repeated doublings cycle between 1 and 2 (mod 3), with the former corresponding with even powers of 2 and the latter corresponding with odd powers of 2. Thus, in this case, 2^xn - 1 is divisible by 3 only if x is even.

Case 2: Since n = 2 (mod 3), we have 2n = 1 (mod 3), 2(2n) = 2 (mod 3), etc. This is similar to Case 1, except in this case, 2^xn - 1 is divisible by 3 only if x is odd.

In Cases 1 and 2, we have shown that another odd natural number can precede n in a Collatz hailstone sequence, whereas in Case 0, n can only be preceded by even natural numbers.

To add another wrinkle, every non-zero terminating decimal actually has two decimal representations, one of which ends in infinitely many 9's (or infinitely many of whatever the highest digit is in other bases). For instance (and most famously), 1 = 0.999.... Other examples include 3.578 = 3.577999, 100.23451 = 100.23450999..., 720934750234567.124 = 720934750234567.123999..., etc. So every number can be expressed as having infinitely many non-zero digits, but certainly not every number is similar to infinity (in fact, no real number is, as others have said).

John, you're taking this thread a bit too seriously, and while there is a point to be made that even joking stereotypes of persecuted minority groups can be harmful, even when made by members of those groups, that hardly applies here.

...unless, of course, you're claiming chemists comprise a persecuted minority group, which might not be too far off the mark given current attitudes and policy relating to science in certain countries.

In any case, this thread is simply a bit of self-deprecating fun, and there's no need to treat it as such serious business.

In practical terms is this change (of an unknown function) computed just by taking the change of the function from the previous iteration? Could it also be done by taking a moving average (exponential smoothing) of the change?

For the first question, if I'm understanding you correctly, then yes. For the second, I don't know.

With regards to the weight adjustment this would mean that if the cost function increases the change (the partial derivative) is positive and so the weight is increased. If the cost function decreases the change is negative and so the weight is decreased. In this way the weight should converge on a value that keeps the cost function at a maximum. If the weight value goes to high and results in a decrease of the cost function the adjustment will be in the opposite direction. (Change signs to minimize rather than maximize the function).

 

Does that sound about right, or are there other things that should be taken into account?

Well, with the caveat that I know very little about machine learning, I believe the idea is to iteratively minimize the cost function. I don't know why the equation on Wikipedia involves addition rather than subtraction. It may be a typo, or it may be that I'm misunderstanding how gradient descent is applied to training deep neural networks.