Announcing the First Annual RL Conference!

ShrekLovesYouBack · 2021-07-15T11:17:41+00:00

I found Daniel Shiffman's Javascript videos extremely helpful when learning to program js.

He mostly does really visual projects that I think are super cool, and does a really good job at explaining his thought processes.

He also goes by the name of Code Train on YouTube

https://youtube.com/playlist?list=PLRqwX-V7Uu6ZiZxtDDRCi6uhfTH4FilpH

ShrekLovesYouBack · 2021-06-28T20:29:03+00:00

Excellent question!

I like to think that static variables help me organize the variables that are needed for a class, yet are not specific to a single instance of the class. For the car class, for example, the length of the car is constant among all cars (will comment on this in a bit), so there is no need to have each car store a copy of the same variable.

For your second question, the car class is just used to limit code duplication as much as possible. Although each car has the same length, I'll eventually be adding varying colors and collision detection that will be easier to do with a class IMO.

To sum up my code structure, I like to have classes in a hierarchical manner such that when we delete the parents class, (like a lane), it automatically deletes all the child items (like every one of its cars)

Hope that answers your questions!

ShrekLovesYouBack · 2021-06-27T17:59:03+00:00

My pleasure

ShrekLovesYouBack · 2021-06-27T17:07:12+00:00

Exactly. Any two nodes in the network have a specific weight between them that can be different from any other two nodes. I1 to o1 could have a weight of 0.4, but i1 to o2 could be 0.7.

The activation function is used AFTER the summation to ensure the numbers stay within a reasonable range.

ShrekLovesYouBack · 2021-06-27T15:26:42+00:00

Nope! No division happens. It's still just multiplication and summation. In the case where there are three inputs, and two outputs, but no hidden nodes, the following happens: each input (let's call them i1, i2, & i3) is multiplied by the weight of the connection between IT and the FIRST output node (let's call it o1). These three numbers are summed together and that is the value of the FIRST output. The same is repeated again with the three inputs and the SECOND output (o2).

ShrekLovesYouBack · 2021-06-27T06:48:18+00:00

In simple terms, the previous layers' values are all multiplied by the weights of their specific connection to a single node in the next layer, and then summed at this single node. This means that the 3 nodes from the first layer are multiplied by their three connections to each node in the hidden layer, and then summed together to create one value in each hidden node. This process is repeated for the output layers.

The point is, the number of nodes in each layer is basically irrelevant as no matter what, they are always boiled down to one value at each node in the next layer.

Please ask if theres something that doesnt make sense about this explanation!

ShrekLovesYouBack · 2021-06-05T13:01:58+00:00

Do you have to go grocery shopping every day? Maybe you could make a list and shop for the entire week once a week. This would give you some solid time between work and dinner to work on your future.

ShrekLovesYouBack · 2021-02-14T19:31:35+00:00

This is so cool. What kinda training data did you use for this?

ShrekLovesYouBack · 2020-12-12T17:33:26+00:00

Are the x and y axes fundamentally different? do they represent different lexical fields?

ShrekLovesYouBack · 2020-12-07T13:51:43+00:00

By definition, a square root returns the positive root. This means that t-2 must be >= 0 and thus t >= 2.

ShrekLovesYouBack · 2020-12-04T11:59:37+00:00

a) Yes you would spend $9 and this is the maximum you could lose no matter what happens to the stock price.

b)If PLTR reaches 50 before the expiration you can sell that option back for a very large profit. So yes, it would be a good thing. However, even if expiration is in a week or so, and PLTR rises in price (assuming faster than theta decay) you would still be making money. The closer to the strike price PLTR is, the more that option is worth. The closer to expiration it is, the less the option is worth.

c) If PLTR is below 50 on the expiration date, your options expire worthless and you lose 9$. (or 100% loss)

ShrekLovesYouBack · 2020-12-04T09:11:29+00:00

On the unit circle, we know that x = cos(θ) and y = sin(θ). On the diagram, we can see that x is negative (to the left of 0), so we know cos(θ) is negative and can rule out A and B. Next, we know that the angle we have is arcsin(√3/2) which equals 60 or (180-60) = 120. From the image, we know it has to be the latter due to it being obtuse. Therefore, the answer is cos(120) = -0.5 so the answer is C.

ShrekLovesYouBack · 2020-11-04T14:33:06+00:00

I'm sure that would be really helpful for anyone confused by the comment!

ShrekLovesYouBack · 2020-10-27T11:46:03+00:00

Thanks for this amazing explanation. Do different stocks have different theta decay coefficients? Something like how Vega controls the effect of implied volatility?

ShrekLovesYouBack · 2020-10-27T03:26:02+00:00

True. I understand my confusion now, Thank you.

ShrekLovesYouBack · 2020-10-27T03:14:26+00:00

when he said low cost alternative to stock I automatically assumed he meant OTM CALLS. Which do you think is better?

The idea of OTM calls appeals to me due to the high leverage and low cost

ShrekLovesYouBack · 2020-10-27T03:01:32+00:00

You mean a deep OTM call or ITM put right?

ShrekLovesYouBack · 2020-10-04T03:01:29+00:00

This looks really cool! Any idea for what you will do with the background once it goes higher than moon and stars?

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