What's the probability of having 3 A's in a row or at least 2 L's, in a string containing only O, A, and L with length m?

Yes, this is PE#191.

Can anybody give me a hint on how to go about doing this? and F_n is the Fibonacci sequence.

As I see it, the main difficulty is with avoiding overlapping subsets. I don't haave a full solution, but I think the best starting split is into...

a) Cases with 0 instances of L
b) Cases with 1 instance of L
c) Cases with 2+ instances of L

C is easily computed as 1-a-b

For cases a and b, you then need to compute the probablity of 3 As, but O and L can just be considered as non-A.

Yeah, my problem is calculating the "3 A's" part.

My current approach is:
AAA(OL)(AOL)(AOL)(AOL)...
(OL)AAA(OL)(AOL)(AOL)...
(AOL)(OL)AAA(OL)(AOL)(AOL)...
(etc)

However, this produces some duplicates.

But for PE, you don't need a formula, you need code...

P3 (nA, nX) => (nA.P2(nA-1,nX)+ nX.P3(nA,nX-1)/(nA+nX)
P2 (nA, nX) => (nA.P1(nA-1,nX)+ nX.P3(nA,nX-1)/(nA+nX)
P1 (nA, nX) => (nA)/(nA+nX)

Working down from the top... The probability of a sequence of 3 breaks down into either an A (of which there are nA remaining), followed by a sequence of 2 (with 1 less A to pick from); OR non-A (nX choices) followed by any string with 3 A's (recursive call, but with one less non-A), all divided by the total number of characters.

Similarly to get a trailing double, you need another A (nA choices) followed by the third (again with one less to pick from). If not, you're washed out and start again looking for a string of 3.

Finally, P1 the probability of an A is just the number or remaining A's over the total remaining characters.


Any help?

Is there a way you would typically do this?

Yea, nevermind I figured that one out. I do have a different question. How many edges does an undirected graph need to guarantee it is connected?

If you have one unconnected node (of N nodes), then all the remaining N-1 nodes could form (N-1)(N-2)/2 connections. Add one more link, and you guarantee that the final node is connected. So I make it 1+((N-1)(N-2)/2).

I hope this isn't homework...?

A fully-connected graph of n nodes has (n-1) * (n-2) / 2 edges. Logically, a graph of n+1 nodes that has 1 node not connected will have at most the same number of nodes.

EDIT: Diav types faster :\

What are determinants used for?

All sorts of things. They can be used for the cross product of two vectors (this also means they can be used to find the curl of a vector field). They can be used to solve systems of equations (via Cramer's rule, but let's face it, Cramer's rule sucks). They are used in finding the inverse of a matrix (which then has all sorts of applications as well). Eigenvalues are the one that I use them for the most (and those can then be used to solve systems of differential equations.

Basically, the list goes on and on.

Maths question:


Volume – the Method of Cylindrical shells
Find the volume of the solid generated when the region y=x/root(x^3+1), the x-axis and the line x=2

Okay, so that’s the question, and I can usually do these sorts of questions. Using the normal volume method of how Volume = 2∏ xy. dx with limits 2 and 0.
So it’s V = 2∏ ∫x²/ √(x³+1)
Using bi-parts integration where,
u =x²
u’ =2x
v = 2(x³+1)^1/2/3x²
v’= (x³+1)^-1/2

By bi-parts, V = 2∏ [2/3(x³+1)^1/2 - ∫4/3x (x³+1)^1/2] with limits 2 and 0.
Then I bring out the 4/3 out of the integration to make life easier, so ignoring the first bit, I do bi-parts again.
V = [...] – 4/3 ∫x^-1(x³+1)^1/2

But then I realise if I use bi-parts again, the cycle will just continue... So what exactly do I do in order to get a simplified integration so that I can substitute 2 and 0 to get the volume of revolution?

Edit: √ is a square root, don't know why it ended up so weird-looking. :/

Wanted to point out an error:

You forgot to square the denominator, so it should be x²/(x³+1)

Use integration by substitution: u=x^3+1, u'=3x^2, ∫x²/(x³+1)=∫((1/3)*1/u)=1/3∫(1/u)

Integrate 1/u and substitute back in the value of u:

1/3*log(u)=log(x^3+1)/3

Should be self-explanatory from this point on:

log(2^3+1)/3-log(0^3+1)/3=log(9)/3-log(1)/3=log(9)/3

Hopefully I did that correctly. It's been a while since I've done volumes of revolution graphs.

Why am I squaring the denominator? I'm not using the V = ∫∏(f(x)²) dx method. This is the shells method.

Here's an example I quickly drew up:
Volume thing.jpg

Oh. Sorry then, I can't help.

WolframAlpha could help, if you need to know how to get the antiderivative.

Ahh, right. Thank you. :)

Ew. Who calls integrals "antiderivatives"?

I refer to them only as "derivative-undoers" and "area-under-the-curve finders."

...

Integrals and antiderivatives are not the same thing, but they are very closely related by the Fundamental Theorem of Calculus. By the FTC, if you can find the antiderivative of a function, you can find the integral of that function over a given interval by subtracting the antiderivative of the lower bound from the antiderivative of the upper bound. Hence beary's suggestion of using Wolfram Alpha to find the antiderivative I suppose.

I'm very glad my major doesn't require me taking any hard math classes. College algebra and I'm done! And I'll probably knock that out as a junior in high school.

Is there an easy way of getting the second derivative of u/v? WolframAlpha says uv''-u(2u'v')+2v(u')²/u³, but I'm sure there's an easy pattern I can follow...

And since we're on the topic of second derivs, (uv)''=u''v+2u'v'+uv''

The quotient rule is just the product rule restated.

Hi, I really don't understand finding components of points on a line, so could someone please just help me with part a) of this question? I'm sure I could finish the question after that, I'm just stuck with finding alpha :(

Since you know that the two lines have to intersect, I would try setting the i and k components of the two lines equal to each other and solving for s and t. You should end up with two equations and two unknown variables, which is solvable. Then equate the j components and use the values for s and t you found to solve for α.

Edit: To clarify, I mean set the i component of the first line to the i component of the second line, and do likewise for the j components.