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**soroban****Member**- Registered: 2007-03-09
- Posts: 452

. . . . . A Mind Reading Trick

1. Choose any two positive integers; write them in a column.

2. Add the two numbers; write the sum on the third line.

3. Add the last two numbers; write the sum on the next time.

4. Repeat step 3 until you have a list of 20 numbers.

5. Select any one of the last five numbers.

6. Divide by the preceding number.

7. Concentrate on that quotient.

. . . . . . Summary

You began with **any** two integers.

. . I have no idea what they were.

You made a list of twenty sums.

. . There is no way I could predict your list.

You selected one of the numbers.

. . Which one? .I don't know.

You divided by the preceding number.

. . No clue that number either.

Yet I can divine your final quotient.

It begins: .One - point - six - one - eight - zero - three . . .

(Are you impressed?)

.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,785

Hi soroban;

Yes, it looks like it is converging to the golden ratio ≈ 1.61803398874...

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**soroban****Member**- Registered: 2007-03-09
- Posts: 452

. .

.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,785

Hi soroban;

Yes, you can use any 2 starting values that are the same and f[n+1] / f[n] will converge to the golden ratio.

*Last edited by bobbym (2009-06-23 06:55:01)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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**Avon****Member**- Registered: 2007-06-28
- Posts: 80

The trick doesn't quite work for *any* two starting numbers. Let

and choose and as the starting numbers.

Since

the sequence we get isso the ratio of two consecutive terms is always

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

The convergence seems to work best if the two positive integers are the same, and worst if the second is zero

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Avon****Member**- Registered: 2007-06-28
- Posts: 80

MathsIsFun wrote:

The convergence seems to work best if the two positive integers are the same, and worst if the second is zero

What do you mean by "work best"?

I can't see a measure of goodness of convergence such that both

i) Starting with 1, 1 works better than starting with 1, 0

ii) Starting with 1, 1 works better than starting with 610, 987

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**MathsIsFun****Administrator**- Registered: 2005-01-21
- Posts: 7,560

I used "seems" as I just played with it in Open Office Calc. My measure of choice was the 9th and 14th terms ... comparing them to a more accurate version of Phi.

1,1: 9th term: 1.61904762

1,0: 9th term: 1.62500000

But you are right ... preloading with terms that are close to the ratio of Phi is better than 1,1.

"The physicists defer only to mathematicians, and the mathematicians defer only to God ..." - Leon M. Lederman

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**Avon****Member**- Registered: 2007-06-28
- Posts: 80

I will admit that my choice of 610, 987 was rather extreme. I certainly wouldn't expect this pair to be found by randomly trying pairs of integers, but even 1, 2 is a better starting pair than 1, 1.

The important thing is that the ratios of consecutive pairs of Fibonacci numbers are the best rational approximations to the golden ratio.

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**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 93,785

Hi;

As soroban pointed out you can start with any constant as the initial values for the fibonacci sequence. For example:

c,c,-> 2c,3c,5c,8c,13c,21c... this sequence is only the fibonacci number times by c, so when you take the ratio (c*F[n+1]) / (c*F[n]) the c's will cancel and you will get F[n+1] / F[n] which converges to the golden mean.

Also when you use other starting values than 1 and 1 you are technically not computing the fibonacci sequence. For example a starting values of 1 and 3

1,3,4,7,11,18,29... this is called the Lucas Sequence.

*Last edited by bobbym (2009-06-23 22:48:18)*

**In mathematics, you don't understand things. You just get used to them.**

**I agree with you regarding the satisfaction and importance of actually computing some numbers. I can't tell you how often I see time and money wasted because someone didn't bother to run the numbers.**

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