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This page has been Tagged As Maths
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Recently I was sent a lovely little twitchet by a friend,
so I thought I'd share it for people to think about. No
answers provided ...
Firstly, let's do something odd ...
Take the powers of 2 up to $2^9$.
- 1, 2, 4, 8, 16, 32, 64, 128, 256, 512.
Now list them in lexicographic order (personally, I find this
a bit tricky!)
- 1, 128, 16, 2, 256, 32, 4, 512, 64, 8.
Now put a decimal point after the first digit:
- 1.0, 1.28, 1.6, 2.0, 2.56, 3.2, 4.0, 5.12, 6.4, 8.0
So that gives us a sequence of numbers in the range from 1 to 8.
Now put that list to one side and ...
... Let's do something completely different ...
n |
Power
|
Ans
|
0 | $10^{0.0}$ | 1.000 |
1 | $10^{0.1}$ | 1.259 |
2 | $10^{0.2}$ | 1.585 |
3 | $10^{0.3}$ | 1.995 |
4 | $10^{0.4}$ | 2.512 |
5 | $10^{0.5}$ | 3.162 |
6 | $10^{0.6}$ | 3.981 |
7 | $10^{0.7}$ | 5.012 |
8 | $10^{0.8}$ | 6.310 |
9 | $10^{0.9}$ | 7.943 |
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Here are some powers of 10. For each number $n$ from 0 to 9
we are computing 10 to the power of $n/10$. This seems to be
unconnected with our previous exercise, I know, but having
done this we can have a look at the numbers we get.
There's no surprise that the answers are in the range from
1 to 8, and they are not especially "nice" numbers, but that
is what you get when you raise a number to a fractional
power (whatever that means).
But looking closely at the last column, with the eye of faith,
you may notice something interesting.
Combining the results ...
Let's augment the table of powers by adding a last column with
the results of the deeply odd process we started with.
n |
Power
|
Ans
| Oddity |
0 | $10^{0.0}$ | 1.000 |
1.00
|
1 | $10^{0.1}$ | 1.259 |
1.28
|
2 | $10^{0.2}$ | 1.585 |
1.60
|
3 | $10^{0.3}$ | 1.995 |
2.00
|
4 | $10^{0.4}$ | 2.512 |
2.56
|
5 | $10^{0.5}$ | 3.162 |
3.20
|
6 | $10^{0.6}$ | 3.981 |
4.00
|
7 | $10^{0.7}$ | 5.012 |
5.12
|
8 | $10^{0.8}$ | 6.310 |
6.40
|
9 | $10^{0.9}$ | 7.943 |
8.00
|
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The agreement really is remarkably good (says he, remarking on
it, and thus making it a self-fulfilling statement). But it
really is remarkably good, and the question one is left with
is:
I know how to do the calculations, I know how to prove that
this is close, but what I'm lacking is a clear, "intuitive"
explanation for people who are not comfortable with detailed
calculation.
Explanations invited.
Send us a comment ...
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