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revised Simplest Solution for a Round Table Q
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revised Simplest Solution for a Round Table Q
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answered Simplest Solution for a Round Table Q
Nov
24
comment How to find the generating function and the closed form for the generating form
... term, $0.$ Another problem is that the index is off by $1.$ So $a_2=-2$ but $(-2)^1=-2$. Similarly $a_3=4$ but $(-2)^2=4.$ So $a_n=(-2)^n$ is not a true statement. To get a correct formula, you need to modify the exponent slightly. With this correction, the formula will be correct for all elements of the sequence except $a_0.$ (It will even work for $a_1$ since $(-2)^0=1.$) You can't really write a simple formula that works also for $a_0,$ so I would just make that a special case. This exception at $a_0$ is easier to deal with in the generating function context.
Nov
24
comment How to find the generating function and the closed form for the generating form
The expression $(-2)^n$ is not the generating function. It is a (not quite correct) formula for the terms in the sequence. A generating function for the sequence $a_0,$ $a_1,$ $a_2,$ $a_3,$ $\ldots$ is $a_0+a_1x+a_2x^2+a_3x^3+\ldots.$ Such an expression with an indeterminate $x$ is known as a formal power series. Your sequence is $a_0=0,$ $a_1=1,$ $a_2=-2,$ $a_3=4,$ $\ldots.$ You need to multiply these numbers by appropriate powers of $x$ and sum the resulting terms to get a generating function. The formula you give does have the problem you mention: it doesn't produce the initial ...
Nov
23
comment How to find the generating function and the closed form for the generating form
Well, $A(x)$ should be in infinite sequence, not polynomial. I can't help much without knowing where you are getting stuck.
Nov
23
comment How to find the generating function and the closed form for the generating form
$-(2^2)=-4,$ and not as $(-2)^2=4.$ Some calculators with unary minus may be exceptions, but it pays to use explicit parentheses when unsure.
Nov
23
comment How to find the generating function and the closed form for the generating form
I think I agree with Tacet that these are the sorts of arithmetic operations that you need to be able to handle correctly with ease before you can tackle a topic like generating functions. Try not using a calculator to compute the powers of $-2.$ A rule you should know is that $a^0=1$ for any $a\ne0.$ (The expression $0^0$ is undefined.) You should also know that whole number exponents mean repeated multiplication. So $(-2)^1=-2,$ $(-2)^2=(-2)(-2)=4,$ $(-2)^3=(-2)(-2)(-2)=-8,$ and so on. You should be careful about parentheses. Expressions like $-2^2$ are interpreted as ...
Nov
23
comment How to find the generating function and the closed form for the generating form
In your post, you wrote the sequence as $$0,\ 1,\ -2,\ 4,\ -8,\ 16,\ -32,\ 64\ldots,$$ which led me to believe that the sequence was infinite. Why do you terminate the sequence with the $x^7$ term? Is it actually a finite sequence?
Nov
22
answered How to find the generating function and the closed form for the generating form
Nov
22
answered Monty Hall Problem with Five Doors
Nov
22
comment Monty Hall Problem with Five Doors
... then you do not get any information about your original selection. But there are small modification of Monty's behavior such that you do gain information. Here's one: suppose that Monty always opens the leftmost unchosen door not containing the prize. Say the doors are labelled A, B, C, D, E from left to right and you pick door A. If Monty responds by opening door B, then your chances of winning by sticking with your original door have increased from 1/5 to 1/4.
Nov
22
comment Monty Hall Problem with Five Doors
I don't believe this argument is right, since it seems to suggest that nothing Monty could ever do involving the other doors could possibly give you information about whether your initial guess was right. What if Monty opened the door with the prize and then asked if you wanted to switch? Or if he opened all of the other doors revealing no prize? In neither case would the probability of winning by sticking remain 1/5. The point of the Monty Hall problem is that if Monty always opens one of the unchosen doors not containing the prize at random, ...
Nov
20
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Nov
18
revised Number of ways to distribute 100 identical chairs among 4 different rooms
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Nov
18
answered Number of ways to distribute 100 identical chairs among 4 different rooms
Nov
18
comment From Icosahedron to Pentagonal hexecontahedron (Floret Tessellation)
According to Mathematica (I used the command PolyhedronData["PentagonalHexecontahedron","VertexCoordinates"]) the vertices are not all equidistant from the center. The 12 original icosahedral vertices are farther from the center than the other 80 vertices in a ratio approximately equal to 1.099457.
Nov
16
comment Flipping coins probability of $6$ flips having more heads than $5$ flips.
... after five tosses then I have a 75% chance of pulling ahead; if we are not tied, then 50% of the time I am ahead (and will stay ahead); if you are ahead by 1, then I pull ahead 25% of the time, and if you are ahead by more than 1, I cannot pull ahead. This makes it clear that the probability I come out ahead is less than 75%, but to compute it exactly, we need to make a detailed calculation like the one in user178543's answer: what is the probability of a tie? what is the probability you are ahead by 1?
Nov
16
comment Flipping coins probability of $6$ flips having more heads than $5$ flips.
An argument very close to the one given by Graham Kemp goes like this: consider the six vs. five case. If we are tied after five tosses then I have a 50% chance of pulling ahead on the sixth toss; if we are not tied after five tosses then there is a 50% chance that I am ahead, in which case the sixth toss doesn't matter, and a 50% chance that you are ahead, in which case the sixth toss still doesn't matter since I can at best equal but no exceed your heads total. Either way, I come out ahead 50% of the time. This argument doesn't work in the seven vs. five case: if we are tied ...
Nov
16
comment Open mathematical questions for which we really, really have no idea what the answer is
... order, that would explain what is known so far, but I have never heard any argument along these lines. From another point of view, one could take the existence of non-Desarguesian planes as reason to be optimistic about existence of planes in non-prime-power orders.