May 07, 2018 Monday
Bedtime Story
Note G of Ada Lovelace - Part 10
Tonight we shall continue with the Note G
of Ada Lovelace, which you need to read with continuous reference to the
flowchart table that I had posted on the night of May 06, 2018 Sunday.
This is a complicated business here that we
are operating upon tonight, so please do not despair if you do not get all the
details.
If you are able to get a grasp of the general
working idea of the engine, consider that as suffice.
“Now Operation 7 must either bring out a
result equal to zero (if n = 1); or a result greater than zero, as in the
present case; and the engine follows the one or the other of the two courses
just explained, contingently on the one or the other result of Operation 7.
In order fully to perceive the necessity of
this experimental observation, it is important to keep in mind what was pointed
out, that we are not treating a perfectly isolated and independent computation,
but one out of a series of antecedent and prospective computations.
Cards 8, 9, 10 produce (-1/2)(2n-1/2n+1) + B1(2n/2) or
.
On Operation 9 we see an example of an
upper index which again becomes a value after having passed from preceding
values to zero.
V11 has successively been 0V11,
1V11, 2V11, 0V11,
3V11; and, from the nature of the office which V11
performs in the calculation, its index will continue to go through further
changes of the same description, which, if examined, will be found to be
regular and periodic.
Card 12 has to perform the same office as
Card 7 did in the preceding section; since, if n had been = 2, the 11th
operation would have completed the computation of B3.
Card 13 to 20 makes A3.
Since A2n-1 always consists of
2n-1 factors, A3 has three factors; and it will be seen that Cards
13, 14, 15, 16 make the second of these factors, and then multiply it with the
first; and that 17, 18, 19, 20 make the third factor, and then multiply this
with the product of the two former factors.
Card 23 has the office of Cards 11 and 7 to
perform, since if n were = 3, the 21st and 22nd
operations would complete the computation of B5.
As our case is B7, the
computation will continue one more stage; and we must now direct attention to
the fact, that in order to compute A7 it is merely necessary
precisely to repeat the group of Operations 13 to 20; and then, in order to
complete the computation of B7, to repeat Operations 21, 22.
It will be perceived that every unit added
to n in B2n-1, entails an additional repetition of operations
(13…23) for the computation of B2n-1.
Not only are all the operations precisely
the same however for every such repetition, but they require to be respectively
supplied with numbers from the very same pairs of columns; with only the one
exception of Operation 21, which will of course need B5 (from V23)
instead of B3 (from V22).”
Stay tuned to the voice of an average story storytelling
chimpanzee or login at http://panarrans.blogspot.com
Good night mon ami and my fellow cousin ape.
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Another great educator and a teacher that I am aware of is
Professor Subhashish Chattopadhyay in Bangalore, India.
While I narrate stories, Professor Subhashish an electronic
engineer and a former professor at BARC, does and teaches real mathematics and
physics.
He started the participation of Indian students at the
International Physics Olympiad.
Do visit him here:
All his books can be downloaded for free through this link:
For edutainment and English education of your children, I
recommend this large collection of Halloween Songs for Kids:
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