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Hi bobbym.
Someone has asked the same question on another forum – I think you can answer it.
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[*]http://www.artofproblemsolving.com/Foru … 6&t=610381[/*]
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Well, let us assume at each step the original line is broken into fragments of equal length. After the first step we have 4 fragments of length 1/4, after the second we have 16 fragments of length 1/16, etc. Then we have
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Hence
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I think you mean dy/dt in the second equation. If you divide the second by the first you should get
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eliminating the parameter t.
There seems to be some confusion here. What Whizzies means is that (excluding complex numbers) log(y) is only defined for y > 0. The problem involves log(x+3), so y = x+3. It is fine for x = −1; it is the y that needs to be > 0.
You’ve got as far as
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Now just solve it for x.
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Why?
It would make no sense for the maximum to be biased towards either x, or y, or z. So It ought to be when they are equal.
That was what I thought as well until I found this:
That might be the maximum but I don’t know how prove it yet (with or without Langrangian multipliers).
I almost got half right.
You did better than over 50% of students from Princeton, Harvard and MIT.
Well, they’re not that difficult.
#1. What is CD in Roman numerals?
#2. The cube of a number is that number multiplied by itself three times – true or false?
#3. What is the pseudonym of the writer Mark Twain?
You’re welcome.
As I said, problems of this kind often demand a lot of ingenuity from the solver, but practice and experience can help a lot. Don’t worry if you can’t come up with such things on the spot – I certainly didn’t for this one! I had to think hard about it for a while. While thinking about it, I realized that the 1 was going to be a problem if I was going to use the triangle inequality, so I tried splitting the 1 into the terms of the sum – and it worked. I mean, it might not have worked and then I would have to start again from the beginning, but fortunately it did. So yeah – practice, practice, practice.
Perhaps those students of Harvard, Princeton and MIT would also be interested in the following questions?
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[*]What is CD in Roman numerals?
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[*]The cube of a number is that number multiplied by itself three times – true or false?
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[*]What is the pseudonym of the writer Mark Twain?[/*]
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If you define N as the minimum of two values, both values must be positive. However for limit problems it’s more usual to define N as the maximum of two values, not minimum.
For problems of this kind you often have to be imaginative. And for this problem I think the key is this:
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Hence:
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Now for r = 1, …, n, we have
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Thus:
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This step
is not valid. Epsilon might be less than or equal to 2.
I don’t really know how you could go about choosing a suitable N for this problem. Let me think about it.
Any plans to make the smilies clickable again?
So the given implication is correct.
So instead of x' you have to type x^{\prime}?
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Off-topic: Why isn’t LaTeX working?