Find matrices A and B if

This just amounts to solving lots of linear equations.

]]>Theorem. For the sequence 1, 2, a3, 4, a5, a6, a7, 8, a3^2, 10, ........

a3 = 3.

Suppose a3 = 3 + h (delta takes too long to keep typing )

So it is sufficient to show that

So just let i get bigger until this is true. This causes a contradiction with the increasing sequence rule, so a3 ≤ 3.

I'll leave showing a3 ≥ 3 as an exercise for the reader

Bob

]]>(x+1)^2 or x^2 +2x +1 as a formula for something. It's obvious to us that they will give the same result even though they look different. Also some methods may seem the same until you input values like 0 or -1/2 into the function so my word description may not work. Anyway does anyone know any resources where I can learn to write word descriptions of patterns in maths notation?

]]>Welcome to the forum.

Sorry, but I don't understand what you mean by " b in degree a "

What branch of mathematics is this from / Whose theorem ? / Can you give an example ?

Bob

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According to your definition, the zero sphere is in R.I'd rather have the zero sphere to be the trivial sphere {0} in {0} and 1-sphere in R.

Well, I guess you'll have to readjust to this definition. It makes more sense when you look at it this way: The 1-sphere is a curved line with only 1 "dimension". The 2-sphere is a surface with 2 "dimensions".

]]>Could I ask, would folks agree with my calculation below:

It is to do with tossing N coins.

By an "experiment" I mean the act of tossing a N coins.

So:

n = Number of experiments performed.

P = Probability of getting N heads in a row after n experiments.

I wish to derive a formula that gives n as a function of P and N.

Here's the reasoning:

On the first experiment, the probability of getting N heads in a row is:

So the probability of not getting N heads in a row is:

If one actually gets N heads in a row at this point then one stops. However, if one does not yet have N heads in a row then one continues.

I'm visualizing this as a probability tree diagram where at each splitting of the tree there are two paths, one with probability

and the other with probabilitySo with this setup, one could ask, what is the probability of obtaining the N heads in a row in two experiments. Well, for this to have happened either "the first experiment would have succeeded" OR "the first experiment would have failed and the second experiment would have succeeded" in producing N heads in a row. So the probability would be:

In this way, we can form an infinite sum which gives us the probability of obtaining the N heads in a row after n experiments, it is:

This is geometric progression and so we can find the sum of the first n terms as:

which simplifies to:

which gives:

So is this correct?

Well we can ask what it gives for the various possibilities below:

1) How many experiments should I perform such that the chances of me getting all heads with 10 coins (N = 10) is 0:

Formula gives

That makes sense.

2) How many experiments should I perform such that the chances of me getting all heads with 10 coins (N = 10) is 1 in 1024:

Formula gives

That makes sense.

3) How many experiments should I perform such that the chances of me getting all heads with 10 coins (N = 10) is 1:

Formula gives

That makes sense.

4) How many experiments should I perform such that the chances of me getting all heads with 10 coins (N = 10) is 0.9999:

Formula gives

(rounded up to nearest whole number of experiments)That makes sense?

My question is, would folks agree with the correctness of this formula?

Thanks,

Mitch.

Bob

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