Re: motivating a kid into analysis

Thanks to James and Dirk for your great posts. Dirk, I'm too busy to
look at yours in detail tonight; I'll give it the attention it deserves
tomorrow. James, it seems to me that what I have to do is choose an n
that corresponds to s(n) = 1/11 or less. If so, the trick is figuring
out the first n that has this characteristic, and then you could choose
that n or any one after that. Would it be 1/(n+2) = 1/11 to find n (n
= 9), and then n = 2^[(9)+1] = 1024? If so, I'm not quite sure how
this relates to epsilon delta.

CS


James Dolan wrote:
> in article <1128533081.319786.271030@xxxxxxxxxxxxxxxxxxxxxxxxxxxx>,
> <zxcv_890@xxxxxxxxxxx> wrote:
>
> |Dirk Van de moortel wrote:
> |
> |> 14 years might be a bit young, but have you tried presenting the
> |> definitions as a real *challenge*? I mean, a function is
> |> continuous in some point if it can withstand the challenge that
> |> whatever number epsilon *he* can come up with, no matter how small,
> |> *you* will always find a delta such that the image of *your*
> |> delta-interval is entirely "projected" within his
> |> epsilon-interval. You can make nice drawings with (not too many)
> |> colours to illustrate this. And make him use a calculator.
> |
> |Hmmm... can you explain what you mean here in a little greater detail
> |(an basic example perhaps?) I've been struggling with the
> |epsilon-delta definition for longer than I care to admit.
>
> before trying to understand the continuity-of-function-f-at-a-point
> challenge game, do you already understand the possibly easier example
> of the convergence-of-sequence-s-to-zero challenge game? in this game
> there are two players, the "epsilon-chooser" and the "n-chooser". the
> epsilon-chooser goes first and chooses an epsilon, and then the
> n-chooser chooses an n. the n-chooser wins _if and only if_ the
> sequence s always stays within distance epsilon from zero once you get
> past the nth term in the sequence.
>
> let's try a couple of rounds of the game; i'll be the epsilon-chooser
> and you be the n-chooser. to make things interesting let's say that
> you gain a dollar every time you win a round and you get hit in the
> head with a shovel every time you lose a round. (you're supposed to
> eventually see that the important question is: do you have some
> perfect strategy for always avoiding getting hit in the head with a
> shovel? that's what it means for the sequence s to converge to zero.)
>
> let's start with the sequence s given as follows:
>
> n s(n)
>
> 1 1
> 2 1/1,000,000
> 3 1/2
> 4 1/2
> 5 1/1,000,000
> 6 1/3
> 7 1/3
> 8 1/3
> 9 1/3
> 10 1/1,000,000
> 11 1/4
> 12 1/4
> 13 1/4
> 14 1/4
> 15 1/4
> 16 1/4
> 17 1/4
> 18 1/4
> 19 1/1,000,000
> 20 1/5
> 21 1/5
> . .
> . .
> . .
>
> (continuing forever following the obvious pattern: 2^n instances of
> 1/(n+1), followed by a single instance of 1/1,000,000, followed by
> 2^[n+1] instances of 1/(n+2), and so forth and so on forever.)
>
> since i'm the epsilon-chooser, i go first, and i choose epsilon :=
> 1/10. now it's the n-chooser's turn to choose. (that's you.) what n
> do you choose?
>
>
> --
>
>
> [e-mail address jdolan@xxxxxxxxxxxx]

.

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