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[Baffler]

I need to "demonstrate how to find the surface area, area, and final position of an arbitrary object moving in four dimensions." There aren't any numbers associated with the problem, he just wants us to show it worked out. I'm really not sure how to approach the problem, but that's my entire test tomorrow.

How to find the surface area depends on how the given object is described. How to find the final position depends on in what form the motion is given. If the question isn't any more specific than what you've quoted, then there is literally nothing more to say. If he wants more details, tell him you too need more details, and you can't read minds.

Although if you're a smartass you could just wait for the object to finish moving and then check its shirt size.

I asked him, and he said that "it's an object in motion, rotating along the x' axis," so I'm thinking that there's just not gonna be any real help on this one ahead of time.

[hops]

Can someone explain to me how to find the integral of 1/(b(x^2)+(a+b)) with regard to x? Assuming a and b are constant and 0>b>a.

[hops]

I wasn't taught about that. Explains why I was unable to solve it then.

[hops]

I'm really having trouble wrapping my head around how to use the Taylor Series. Can someone explain it to me in layman's terms?

[Helgoland]

Functions in general are way complicated. Polynomials on the other hand are fairly simple and well-understood. So it would be pretty cool if we could write every function as a polynomial, right?

Taylor series try to do this step-wise. First you pick a point that you're interested in and take a very simple polynomial that looks like your function at that point. Then, to correct for this very simple polynomial becoming very dissimilar to your function further away from that point, you make it slightly more complicated. Then you correct again, and again, and again... And every time your approximation becomes better. After 'infinitely many corrections', it becomes perfect! The practical value of Taylor Series though is that a fairly rough approximation is usually good enough.

A valuable example is doing the Taylor Series of a polynomial. Yes, it sounds stupid, because we'll just get the original polynomial back, but maybe it'll help you grasp the concept. (I'll skip the steps where I calculate the coefficients, that you'll have to do yourself.)

We want to approximate, let's say, x^4 + 3x^3 - x^2 - 10x + 3, at the point 0.

First we take a very simple approximation: 3. Works fine at 0, sucks pretty much everywhere else.
Refining this, we get -10x + 3. Better, but not great.
Then we get -x^2 - 10x + 3. Hey, it's starting to make wobbles and everything - we're getting somewhere!
Then comes 3x^3 - x^2 - 10x + 3. Aaaaalmost good, but still weird far away from our point of interest 0.
The next approximation, x^4 + 3x^3 - x^2 - 10x + 3, is pretty good though!

Try doing this explicitly - calculating the coefficients, not just reading them off! - for, I dunno, x^3 - 4x^2 + 2x + 1.

[TheDarkStar]

To explain a bit more, basically what you do is find a function whose value and first, second, third, etc. derivatives equal the value of your function at a specific point. At the point you chose, the two functions should be exactly equivalent up to whatever derivative you're using. As you get further and further away, the functions will diverge, but they will stay relatively close as long as you pick points close to the original.

[Arx]

In case you meant where you might actually need it, it's sometimes (often) incredibly convenient to take something like log(x-n) and turn it into a polynomial instead.

[hops]

Man, I should've made an attempt to understand this way earlier in the semester.

[Andres]

What are the odds of double sixes when rolling 4d6? How do you work it out?

[Reelya]

The general solution is to use the binomial distribution.




k is the number of successes you're testing for (2), n is the number of dice (4), p is the probability of success (1/6).

As for an explanation, note that there is a term p^k, which is the probability of 2 successes, and it's multiplied by (1-p)^n-k, which is the probability of two failures: which means you're measuring the chance of getting two 6s followed by 2 non-6s. But since we don't care what order the 6s are in, we can work out how many possible ways 2 things could be arranged in 4 slots (equation 2, aka the binomial coefficient), then since each of those patterns is just as likely as the first one we worked out, we can multiply the chance of that one pattern by the number of patterns to get the total chance of gettings 2 sixes in 4 dice.

[hops]

Here's an inverted version for Darkling users.

[Andres]

The general solution is to use the binomial distribution.




k is the number of successes you're testing for (2), n is the number of dice (4), p is the probability of success (1/6).

As for an explanation, note that there is a term p^k, which is the probability of 2 successes, and it's multiplied by (1-p)^n-k, which is the probability of two failures: which means you're measuring the chance of getting two 6s followed by 2 non-6s. But since we don't care what order the 6s are in, we can work out how many possible ways 2 things could be arranged in 4 slots (equation 2, aka the binomial coefficient), then since each of those patterns is just as likely as the first one we worked out, we can multiply the chance of that one pattern by the number of patterns to get the total chance of gettings 2 sixes in 4 dice.

How do I put this into a calculator? The scientific one that Windows 10 comes with, specifically.

Also, is mc² (m)*(c²) or (mc)²?

[i2amroy]

How do I put this into a calculator? The scientific one that Windows 10 comes with, specifically.

Also, is mc² (m)*(c²) or (mc)²?

The Windows 10 calculator has a factorial function yeah? So just... plug it in. (If you're really having issues with it then just throw it in wolfram alpha instead, but you should just be able to substitute into the final equation and plug stuff in just fine).

And if you're talking about e=mc² then it's m*c².

[Andres]

0.16. With the two equations (minus the end bit of the first since I'm not looking for non-6's after I get two) I get 0.16. So there's a 16% chance that rolling four 6-sided dice will end up with two of them rolling 6's.

What's the equation for finding the odds of rolling at least two 6's on four 6-sided dice?

[MrRoboto75]

What's the equation for finding the odds of rolling at least two 6's on four 6-sided dice?

You could just run the equation with k equaling 2, 3, and 4 and just add those three chances together.