[Potironette]

So..the dark blue I see at night is actually bits of blue light arriving from the other side of the planet :o?

Also, just a sort of random question out of curiosity: when a light in a room turns on, does the light actually travel/move to fill up the space or does everything just magically light up at once? Or, if I have a laser on top of one hill hitting a mirror over 186,282 miles away angled to hit some surface beside me, will there be a delay in my seeing that surface? Actually, this is probably a badly thought out question and stuff will take the time of light speed--and it just looks like a dark room fills up with light at once because light speed is so fast, and rooms so small.


I have a math review problem (midterms are coming up) that asks me to find a factor of (j-3)^2 - (k-5)^2 but I'm not sure how to do it. I tried to expand it, but just got j^2 - 6j - 16 - k^2 + 10k and it doesn't seem useful. Should I have expanded it at all?

[Coda]

Originally Posted by Potironette

So..the dark blue I see at night is actually bits of blue light arriving from the other side of the planet :o?

Yes, exactly. If the atmosphere wasn't there, it would be completely black. Pictures taken from the moon support this.

Also, just a sort of random question out of curiosity: when a light in a room turns on, does the light actually travel/move to fill up the space or does everything just magically light up at once?

Neither! Light doesn't fill space. When you turn the light on, the bulb begins to emit photons. Those photons bounce off of the things in the room and you see the ones that end up ricocheting into your eyes.

But yes, objects closer to the light source will be hit with photons before objects farther away from it.

Or, if I have a laser on top of one hill hitting a mirror over 186,282 miles away angled to hit some surface beside me, will there be a delay in my seeing that surface?

This experiment has actually been carried out for real!

When they visited the moon, astronauts left behind a retroreflector on the surface. (Retroreflectors are the things like on street signs and bicycles that are textured so that no matter what angle incoming light hits the surface, at least some of it is sure to bounce towards your eyes.) If you have a powerful enough laser (it has to be powerful because the atmosphere is going to make it spread out and lose energy), and you shine it at that retroreflector, you won't see it light up for about 2.5 seconds. (Actually, you won't SEE it light up at all because the reflection is so weak by the time it gets back here, but it's measurable by instruments.)

This technique has given us a VERY precise figure on EXACTLY how far the moon is from the earth -- and how fast it's spiraling away from the planet (about 3.8cm/year).

Actually, this is probably a badly thought out question and stuff will take the time of light speed--and it just looks like a dark room fills up with light at once because light speed is so fast, and rooms so small.

No such thing as a bad question. It shows you're exploring the problem space and seeing what else there is to be found.

Though regarding the dark room, light speed isn't actually the only factor involved. The bulb itself takes time to warm up and start emitting light in the first place. So if it feels instant between hitting the switch and seeing things, that's just a sign of how bad humans are at perceiving very small intervals of time.

I have a math review problem (midterms are coming up) that asks me to find a factor of (j-3)^2 - (k-5)^2 but I'm not sure how to do it. I tried to expand it, but just got j^2 - 6j - 16 - k^2 + 10k and it doesn't seem useful. Should I have expanded it at all?

Nope, you shouldn't have expanded it. It's a formula you should have memorized already: a^2 - b^2 = (a + b)(a - b)

[Tiva]

So Coda is imparting his worldly knowledge on everyone here?

I can help with Physics and Anthropology… if I remember to check the thread.

[Potironette]

Ohh I completely forgot about a^2 - b^2 = (a+b)(a-b). Thank-you!
So (j-3)^2 - (k-5)^2 --> (j - 3 + k - 5)(j - 3 - k + 5) --> (j + k - 8)(j - k + 2) and these two are the factors!

Wow, it's amazing it's actually been done :o. I based that question off of some telegraph experiment I learned about last year, which I don't remember too clearly.
But, uh, the moon is spiraling away from the Earth..? Granted it's not a lot every year but if Earth survived long enough it's just going to spiral away?

Light bulbs emit photons and those go into my eyes so..the fact everyone in a room sees the whole room light up for an instant when the switch is flipped on and off is because light moves and bounces quickly?

[Coda]

Originally Posted by Tiva

So Coda is imparting his worldly knowledge on everyone here?

I can help with Physics and Anthropology… if I remember to check the thread.

You're welcome to contribute!

Originally Posted by Potironette

Ohh I completely forgot about a^2 - b^2 = (a+b)(a-b). Thank-you!
So (j-3)^2 - (k-5)^2 --> (j - 3 + k - 5)(j - 3 - k + 5) --> (j + k - 8)(j - k + 2) and these two are the factors!

Wow, it's amazing it's actually been done :o. I based that question off of some telegraph experiment I learned about last year, which I don't remember too clearly.
But, uh, the moon is spiraling away from the Earth..? Granted it's not a lot every year but if Earth survived long enough it's just going to spiral away?

Light bulbs emit photons and those go into my eyes so..the fact everyone in a room sees the whole room light up for an instant when the switch is flipped on and off is because light moves and bounces quickly?

Yes, yes, yes, and yes.

[Tiva]

The other reason that everyone see the bright flash of light when you turn to bulb on is because the light travels faster than your eyes can adjust to it. So the brightness is caused by the delay of your pupils adjusting from the light meeting your eyes. Also why most people blink when they walk outside or see a bright light, to protect your pupils while they adjust down to let in less light.

[Potironette]

Is the light that I see not the same light that another person sees?

The pupils take time to adjust...do you mean that when the light first turns on and off, people don't see the room and just see the bright flash and that's because the eye can't make out the details because it needs to adjust? Why do pupils need to be protected :o?
I mean, it makes sense pupils need to be protected, but why is light harmful? Is it like the sunspots thing where when I look at a super bright object I see a blob in my eyes for a while? What is that?

[Coda]

Originally Posted by Potironette

Is the light that I see not the same light that another person sees?

Nope. It comes from the same source, but the photons are distinct.

The pupils take time to adjust...do you mean that when the light first turns on and off, people don't see the room and just see the bright flash and that's because the eye can't make out the details because it needs to adjust?

Yes. A less-bright pulse of light, on the other hand, can do a fairly good job of giving you an impression of the room, but your body's reaction will make the darkness afterward feel even darker.

This is why you should try to avert your eyes from oncoming headlights when driving at night, by the way -- the bright light in your face diminishes your low-light vision.

Why do pupils need to be protected :o?
I mean, it makes sense pupils need to be protected, but why is light harmful? Is it like the sunspots thing where when I look at a super bright object I see a blob in my eyes for a while? What is that?

That's retina burn. Mild cases just result in the nerves in your eyes getting overloaded for a while and sending off bad signals until they recover -- that's what the spots are. More serious cases, like looking into the sun for more than a brief moment, can lead to permanent blindness. This risk of damage is why we have reflexes to make us blink and look away, because we as a species depend so strongly on our sense of vision.

EDIT: Forgot to explain why light is harmful. Long story short, light is energy. Put too much energy into a living cell and it cooks. Your eye has a lens in it that concentrates light onto the retina, and your retina is made of very very sensitive cells, so it takes less light to cause damage there than to -- for example -- your skin. (On the other hand, the lens filters out the ultraviolet light that can cause DNA damage, so unless you've had cataract surgery or something that removes the lenses you don't have to worry about sun-induced retinal cancer like you have to worry about skin cancer.)

[Potironette]

Ah, so that's why everyone can see in a room XD. Actually, when I think about it, once light hits someone's eyes, doesn't it go bouncing off (or probably re-emitted would be better) as the color of the eyes?

Ultraviolet light can cause DNA damage? I remember reading a long time ago about some person who was making pictures of how he was seeing UV light after surgery (I searched it up again: http://www.komar.org/faq/colorado-ca...et-color-glow/). So...any time he's seeing this UV light he's at risk for cancer from DNA damage? And in fact anyone with cataract surgery risks that?

And thanks Tiva and Coda about the eyes!

[Tiva]

Do you mess around with Cameras? It is the same idea as your eyes. You have a lens that collects light to process an image. Too much light and it over saturates the image like a photo taken in high light. Too little and the image isn't sharp and colors less pronounced. When you turn on the light your eyes take a few seconds to process and close the aperture that lets in light aka your pupil.

[Potironette]

I've only slightly messed around with one camera ^^;;. Buut even though I don't know what an aperture is, I can understand that at first, the eyes are adjusted to darker lights by having the pupil let in more light... Ohh! Then suddenly the light turns on and now there's too much light to see, for which the pupil needs to take time to adjust for because it's a body part, and then the light goes away before the pupil adjusts. And that's why people only see a flash and not the objects in the room!

Does that mean a less bright pulse of light gives an impression of the room because it requires less time for the eyes to adjust? Or that the eyes don't really need to adjust much but it does anyway and that's why it's harder to adjust back to less light when the pulse of light goes away?

[Coda]

You're basically right on all counts!

[Potironette]

What is a frequency in graphing? And can tangent, cotangent, secant, or cosecant graphs have frequency? All I know is, if I need to answer what a frequency is on my test, I'm going to write the reciprocal of the period. But I don't actually know what the frequency is.


Also, why do arccos, arcsin, and arctan exist as functions? It's weird that arctan is the only one that extends forever along the x-coordinate. Whereas stuff like arccos(2) or arcsin(2) turns out undefined.

[Coda]

The frequency is how frequently (hence the name) a periodic function (such as sin) goes through its full cycle per unit time. That's why it's 1 over the period -- one unit of time, divided by how many units of time must elapse for a single cycle, is clearly equal to how many cycles are in a single unit of time.

arcsin, arccos, and arctan exist because they're inverse functions. If you know that sin x = some value, but you don't know what x is, then arcsin allows you to find that. And a property of an inverse function is that, over its domain and range, f(f'(x)) = f'(f(x)) = x. For example, if f(x) = 2x, then f'(x) = x/2, because f(f'(x)) = f'(f(x)) = 2(x/2) = (2x)/2 = x.

As for why arctan is the only one that extends forever over the x axis... well, like I said, these are inverse functions. Plot x = sin(y), x = cos(y), and x = tan(y) on a graph, but limit them to a single cycle of the functions, and you'll see that the graph of x = sin(y) is exactly the graph of y = arcsin(x), the graph of x = cos(y) is exactly the graph of y = arccos(x), and the graph of x = tan(y) is exactly the graph of y = arctan(x).

So only arctan goes on forever in both horizontal directions because only tan goes on forever in both vertical directions -- sin and cos only have values between -1 and +1.

These functions are ridiculously useful in real-world practice. I use arctan a lot.

Think about a right triangle with one point at the origin, the right angle on the x axis, and the third point at (x, y). Call the angle at the origin θ. Since it's a right triangle, you know SOH-CAH-TOA. And since tangent = opposite / adjacent, then you know that tan θ = y / x.

Now get rid of the triangle.

You've got an arbitrary point (x, y) in 2D space now. Draw a ray from the origin through that point. What angle does that ray make with the x axis?

tan θ = y / x
arctan tan θ = arctan (y / x)
θ = arctan(y / x)

Now, technically, this is ambiguous, because tan's range is only (-pi/2,pi/2], not (-pi,pi], so you have to look at the signs of x and y to determine which quadrant θ is actually in, but that's easy to do.

So in other words, arctan lets you figure out what angle you have to turn in order to face a given point. You can see why this would be SUPER useful in video games.

[Potironette]

So..arcsin and arccos need to be functions because functions are more useful than relations (I think that's what they're called?) since those give just one output. Or maybe when a function gets inversed(?) people like them to remain functions and so limit them..? That doesn't really seem it though.
(if I just plot x = sin(y) or x = cos(y) I get a relation on desmos)
On the other hand, arctan can go on forever in the x direction and still give one output/be a function which is why they get to go on forever?

The bit about the video games was really interesting :o. Although, though I have a sense that angles are useful in video games I don't really know why figuring out an angle based on a coordinate would be useful.


I found this in my review sheet but I have no clue what it means:
Regular division: p = qd + r, r < d (dividend = quotient x divisor + remainder)
p(x) = q(x)d(x) + r(x) deg(r) < deg(d)

Taking d(x) = x - x_0,
p(x_0) = 0 <-> r(x_0) = 0 <-> p(x) = (x - x_0)q(x), i.e., (x - x_0)|p(x)

(It's the "Taking" part I don't understand. I'm pretty sure deg(r) < deg(d) is something about how remainders don't have the same "degree" as divisors(?), though I don't really know what a "degree" is. I used "<->" to replace the arrows in my review sheet. I'm not sure what "|" is supposed to be, though I vaguely remember reading somewhere it means "divides")


EDIT: Below those equations on dividing, the Remainder Theorem, Factor Theorem, Fundamental Theorem of Arithmetic, Fundamental Theorem of Algebra, and Rational Root Theorem are listed out...and they all refer to p(x) = (x - x_0)q(x). I guess it means to say that somehow, x - x_0 got rid of the remainder, somehow.?

[Coda]

Originally Posted by Potironette

Or maybe when a function gets inversed(?) people like them to remain functions and so limit them..? That doesn't really seem it though.

Actually, that IS it. For the same reason, square root is constrained to only the POSITIVE square root of the number, even though its inverse operation, x^2, exists for all x.

(if I just plot x = sin(y) or x = cos(y) I get a relation on desmos)

I'm not sure what you mean by a relation in this context. Usually when I think of relations I'm thinking of sets.

On the other hand, arctan can go on forever in the x direction and still give one output/be a function which is why they get to go on forever?

Yes. Like I said, it's because tangent goes to infinity in both directions between -pi/2 and +pi/2.

The bit about the video games was really interesting :o. Although, though I have a sense that angles are useful in video games I don't really know why figuring out an angle based on a coordinate would be useful.

You locate objects in space by their coordinates. A simple example would be a 2D game that counts pixels as the coordinate system. If you have a bird enemy that can dive-bomb the hero, then it might be flying along at y=600 while the hero is running on the ground at y=0, and when it gets close enough to see, it needs to figure out what angle to dive at in order to attack.

I found this in my review sheet but I have no clue what it means:
Regular division: p = qd + r, r < d (dividend = quotient x divisor + remainder)
p(x) = q(x)d(x) + r(x) deg(r) < deg(d)

Taking d(x) = x - x_0,
p(x_0) = 0 <-> r(x_0) = 0 <-> p(x) = (x - x_0)q(x), i.e., (x - x_0)|p(x)

(It's the "Taking" part I don't understand. I'm pretty sure deg(r) < deg(d) is something about how remainders don't have the same "degree" as divisors(?), though I don't really know what a "degree" is. I used "<->" to replace the arrows in my review sheet. I'm not sure what "|" is supposed to be, though I vaguely remember reading somewhere it means "divides")

I had to look this up because I couldn't figure out the context: This is specifically talking about polynomial division.

The degree of a polynomial is the highest exponent in it -- the degree of "x + 1" is 1, the degree of "x^2 - x + 1" is 2, the degree of "x^3" is 3, etc.

The degree of the remainder in a polynomial division MUST be less than the degree of the divisor. Think about it by analogy with regular division -- regular division is like repeated subtraction, and if the remainder is larger than the divisor, then your quotient isn't big enough and there's still room to subtract more divisors from the dividend.

The double-headed arrow means "if and only if", and it means that if the left side is true then the right side is also true, and if the right side is true then the left side is also true. For example, "2x = x <-> x = 0" says "if 2x = x, then x = 0, and if x = 0, then 2x = x". On the other hand, "x = 2 <-> x^2 = 4" is invalid, because while "if x = 2 then x^2 = 4" is true, "if x^2 = 4 then x = 2" is false because x could also be -2. (You would just write "x = 2 -> x^2 = 4"; the single arrow is read "implies".)

I'm not super familiar with the property as it's written, but some searching around shows me that this is a property of a polynomial zero. From there, I understand a lot more about it.

"Taking" there could also be interpreted as "suppose" -- that is, suppose you have d(x) = x - x_0. The following must either all be true, or all be false:

1. p(x_0) = 0 -- that is, the original polynomial crosses the x axis at x_0, or equivalently, x_0 is a root of p(x)
2. r(x_0) = 0 -- that is, the remainder of p(x) / (x - x_0) is 0 at x_0
3. There exists some q(x) such that (x - x_0) * q(x) = p(x) -- equivalently, (x - x_0) is a factor of p(x).

This is a formal way of saying "(x - x_0) is a factor of p(x), if and only if p(x_0) = 0".

For example: We can show that the function (x^2 - x) = 0 at x = 1. This property lets us immediately know that (x^2 - x) is divisible by (x - 1). It doesn't tell us what the result of the division is, but we know that a result exists. But it equivalently means that if we know a degree-1 factor of a polynomial, we know that it must touch the x axis at that point.

EDIT: Below those equations on dividing, the Remainder Theorem, Factor Theorem, Fundamental Theorem of Arithmetic, Fundamental Theorem of Algebra, and Rational Root Theorem are listed out...and they all refer to p(x) = (x - x_0)q(x). I guess it means to say that somehow, x - x_0 got rid of the remainder, somehow.?

It means that all of those theorems only hold if p(x) is evenly divisible by (x - x_0). If it's NOT evenly divisible (that is, if r(x) isn't zero) then the theorems don't apply.