Hello good friend George,

Magnificnent!

Willem Bouman



----- Oorspronkelijk bericht -----
Van: george972453@...
Aan: Mental Calculations
Verzonden: maandag 22 juni 2009 20:25
Onderwerp: [Mental Calculation] Boot That Root! (second attempt - using the
other boot?)







Hi folks, I'm back - just like the proverbial bad penny. However, I hope I can
be a bit more useful than a duff coin!

I've been working on a system for determining deep roots with reasonable
accuracy, and I think I'm on to something. Please read on, and please do let me
know what you think:

High powers of less-than-user-friendly numbers can really help to increase the
accuracy of the extraction of a deep root. However, they're also very difficult
to find... aren't they?

Well, no - they're not. As long as you're not too unfamiliar with powers of
single-digit numbers (such as 7^5 = 16807, for example), you can combine these
with a relatively simple mathematical technique to generate the same powers of
non-integer numbers. This means you can get much closer to the actual value with
your estimates, thus gaining more accurate evaluations of how quickly values
will change, and so achieve a final answer with greatly enhanced accuracy.

Consider the problem of obtaining the fifth root of the number 44000. For this
exercise, I shall be using the method I have described in my book - and I shall
be working to six decimal places.

Firstly, we can see that the root is somewhere between 8 & 9, and that it
seems to be a little closer to 8 - so that's our starting point.

8^5 = 32768, and 44000-32768 = 11232 (this is a shortfall).

The gradient of the graph of y=x^5 at x=8 is 5x8^4 = 20480.
The rate of change of gradient is 5x4x8^3 = 10240.

Comparing the shortfall with the current gradient, we see 11232/20480 =
0.548438. Since the average gradient may be assumed to be around half way
ebtween the initial and final gradients, we add half this figure, multiplied by
the rate of change, tothe current gradient - and we find that
20480+(0.548438/2x10240) = 23288.00256. Since this is therefore our estimate of
the average gradient between x=8 and x=(8+0.548438), we multiply by that
interval to give us 23288.00256x0.548438 = 12772.025548.

The shortfall from our original estimate of x=8 was 11,232. Therefore we have
'overshot the mark' by 12772.025548-11232 = 1540.025548.

The gradient at x=8.548438 is 20480+(0.548438x10240) = 26096.00512.

Since 1540.025548/26096.00512 = 0.059014, we drop back this far from our
secondary estimate. 8.548438-0.059014 = 8.489424, and this is thus our final
assessment of the fifth root of 44000, With the correct answer being 8.485756,
the error is 0.003668, and our answer is correct to two decimal places.

Let's sharpen that up a bit. (ok, let's sharpen it up a lot!)

If only it wasn't such a nasty task to find 8.5^5, or even 8.5^4 or just
8.5^3, we could get a better starting position. Isn't there a way we could
improve things?

Well, you ought to have guessed by now that the answer is 'yes'. It all makes
use of the expansion of the function (x+y)^n, which begins with x^n + nx^(n-1)y
+ n(n-1)/2(x^(n-2))y^2.... and eventually ending with y^n. The coefficients of
the individual terms of this expansion are given by the following simple
triangle:

1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
1 6 15 20 15 6 1

Each number, as you can see, is the sum of the two numbers most immediately
above it.

By adjusting the magnitude of 8.5, i.e. moving the decimal point one space to
the right, we make it 85 (which is, of course, 80+5). This means we can now
treat 8.5^5 as 85^5, using the expansion of (x+y)^n, where x=80, y=5, and n=5 -
just as long as we remember to scale down the answer accordingly (by a factor of
100,000).

We see, from this, that 85^5 = 80^5 + (5x80^4x5) + (10x80^3x5^2) +
(10x80^2x5^3) + (5x80x5^4) + 5^5. Ok, so the number 80 is a big one -
apparently. But it's really just an '8' in disguise, multiplied by a factor of
10.

Now we get 85^5 = 3276800000 + 1024000000 + 128000000 + 8000000 + 250000 +
3125 = 4437053125, and so after the reduction by a factor of 100000, we have a
final result of 8.5^5 = 44370.53125, which is correct.

By using the same style of process to obtain 8.5^4 (5220.0625) and 8.5^3
(614.125) for use in calculating the gradient & rate of change of y=x^5 at the
point where x=8.5, we find a gradient of 26100.3125 and a rate of change of
12282.5.

Since we now have an excess value of y by 370.53125, we need to reduce the
value of x by approximately 370.53125/26100.3125 (excess over gradient) which is
0.014196. This gives us x = 8.5-0.014196 = 8.485804. Using the rate of change of
gradient of 12282.5, we can see that the gradient decreases by 174.362 between
x=8.5 and x=8.485804. The average gradient will therefore be reduced by half
this amount (i.e. 87.181) and would thus be 26100.31-87.181 = 26013.129. The
value of 'y' would thus be reduced by 26013.129x0.014196 = 369.282379. This
amount is 370.53125-369.282379 = 1.248871 short of the reduction we require.

The gradient at x=8.485804 is 26100.31-174.362 = 25925.948, and so we need to
reduce the value of x by a further amount of 1.248871/25925.948 = 0.000048.

Since 8.485804-0.000048 = 8.485756, we now have our final answer - which is
correct to six decimal places.

Without going into the maths of the system again, the accuracy of this method
can be tested against the determination of the fifth root of 50000. Working from
the fifth power of integers only for the initial estimate, we reach a final
figure of 8.704887 which is correct to two decimal places & bears an error of
0.000619.

Using the fifth power of 8.7 to grant us our first try, we finally arrive at a
value of 8.705506 which, like before, is actually correct to six decimal places.

Any comments, suggestions, etc. would be gratefully received, either via the
group e-mail system or by my personal e-mail which you should all now have.

Regards,

George Lane

[Non-text portions of this message have been removed]





[Non-text portions of this message have been removed]