answer a financial math question

[KrazyHorse]

flubber, it would help if you would pay attention. Nobody is claiming that under all possible movements in prices the person who exercises later will earn a higher payoff. What I'm explaining to you is that the MARKET PRICE OF THE OPTION is always >= its exercise value. In your example, I guarantee you that the price of the option at the umderlier's peak was greater than 41$.

You appear to lack any understanding of the relationship between underlying and derivatives prices imposed by arbitrage.

[Flubber]

flubber, it would help if you would pay attention. Nobody is claiming that under all possible movements in prices the person who exercises later will earn a higher payoff. What I'm explaining to you is that the MARKET PRICE OF THE OPTION is always >= its exercise value. In your example, I guarantee you that the price of the option at the umderlier's peak was greater than 41$.

If you would have said earlier that it was sub-optimal to exercise early since there would always be more value in selling the option (just another form of cashing in really!!), I would not have had to follow up. It was just that your wording was ambiguous and since I had never spoken about trading the option (merely talking about average folks and exercise) I thought that perhaps there was some mathematical magic involved.

You appear to lack any understanding of the relationship between underlying and derivatives prices imposed by arbitrage.

You mean like your gem that the option value will be at least the exercise value?? Well gollee gee Elmer-- I got this here doodad that someone will give me 41 bucks fer right now or later they might give me more or they might give me less--

Sorry to be sarcastic but I REALLY thought you were trying to go on about something that might be hard to understand. (And just so you understand the options in my example were employee options which could be cashed through the TSAR option without transaction fees and which are not readily tradable-- so when I was talking about best time to exercise I was really talking about best time to exercise-- )


Since I am genuinely interested in some of this stuff I did read a little financial math type stuff last night-- I just flipped through a text available online. The concepts aren't that difficult once you get past some of the jargon. I'll read more. If you ever thought to 1) assume intelligence and 2) maybe attempt to increase understanding with respect to a topic you studied and others have not instead of constantly needing to prove how smart or informed you are (What is up with that ???) you could be a great resource

[KrazyHorse]

If you would have said earlier that it was sub-optimal to exercise early since there would always be more value in selling the option (just another form of cashing in really!!)

/

This post is utterly ridiculous. I suggest you read a basic financial math book. Real investors, if they wish to take a market view, will resell the option for >= the intrinsic value in the market (strictly greater if implied volatility is anything other than 0). If the price was ever less than the intrinsic value then there exists a simple arbitrage. Exercising early is always suboptimal, modulo transaction costs and some minor difficulties like shorting at a retail level.

Like I said, please pay attention

[KrazyHorse]

And no, selling and exercising are not just different forms of cashing in.

The point is that EVEN IF YOU'RE PREVENTED FROM EXERCISING EARLY the option price is higher than or equal to the intrinsic.

[KrazyHorse]

And no, selling and exercising are not just different forms of cashing in.

The point is that EVEN IF YOU'RE PREVENTED FROM EXERCISING EARLY the option price is higher than or equal to the intrinsic.

[Ben Kenobi]

Ok.

Block exerts a force = F= ma which basically acts to change the mass of the overall system from M to 3M.

So the coefficient of static friction now is 3M(mu) as opposed to M(mu), meaning that the Frictional force is 3x the original frictional force.

F= Fm - Ff =

v= S

Ek = 1/2 mS^2.

Ef = Eo

1/2 mS^2 = 1/2 3mT^2

sqrt(S^2/2m) = T

so that will give you the speed of the combined blocks.

For question one.

Vi = 0

Vf = sqrt(S^2/2m)

vf= vi + a*t

sqrt(S^2/2m) = g*t

t = sqrt(S^2/2)/g

[Ben Kenobi]

HC, maybe you can answer a question that Kuci got wrong.

Betelgeuse is a star located 640 ly away from the earth.

RA is 5h 55m

Dec is +7 deg 24'

What are the coordinates of betelgeuse in cartesian coordinates, assuming the Sun is 0/0/0

[N35t0r]

A long uniform slab of with mass M is sliding on a "frictionless" horizontal surface

So the coefficient of static friction now is 3M(mu) as opposed to M(mu), meaning that the Frictional force is 3x the original frictional force.

0/10

On two accounts. First, there's friction between the block and the slab, not the slab and the surface. The force equals Mu*N = Mu*2M*g. Even if there were friction between the slab and the surface, the coefficient doesn't change. The force changes.

[giblets]

I don't know much physics, but... wouldn't momentum be conserved? So tripling the mass of the system causes the final velocity to be one third of the original.

maybe Ben could explain this stuff:

Vi = 0

Vf = sqrt(S^2/2m)

vf= vi + a*t

sqrt(S^2/2m) = g*t

t = sqrt(S^2/2)/g

???

[Ben Kenobi]

That's how I solved for the velocity. Energy is conserved. You can use momentum too.

The rest is kinamatics equation. I know the final speed, and the initial speed, the acceleration and from that I can find the time.

[Ben Kenobi]

The force equals Mu*N = Mu*2M*g. Even if there were friction between the slab and the surface, the coefficient doesn't change. The force changes.

Ok. I missed the part where it said the connection between the ground and the slab was frictionless. I thought they were asking for that, not for the friction between the slab and the block.

[KrazyHorse]

This is so sad...

[Hauldren Collider]

this is ****ing hilarious what are you talking about krazyhorse

[Ben Kenobi]

Obviously.

Sucks when I'm right.

[Hauldren Collider]

That's how I solved for the velocity. Energy is conserved. You can use momentum too.

The rest is kinamatics equation. I know the final speed, and the initial speed, the acceleration and from that I can find the time.

ROFL
ROFLROFLROFLROFL
I AM RIIIIIIDING THE ROFLCOPTER
BAHAHAHA

It's kinetic friction you dumb ass, we aren't accounting for the waste heat so in this system we don't conserve energy. Jesus christ.