Summary:
Count + + Numbering system
+ Addition Stealing and
- Multiplication
Roman numerals
The Abaco (or bullets)
The method of "Prostaferesi "
I" rods Napier "
Division
+ + + Square Root
Exponentiation
+ Logarithms
+ slide rule
Multiplication is a simplified way to calculate a sum of many terms equal, is a task that can be done simply by counting (as explained here ) or, indeed, adding up repeatedly.
The multiplication problems does not subtraction, which also brings up the zero and negative numbers: the multiplication of two natural numbers always gives a natural number (the system of natural numbers is said to is closed with respect to transactions propagation, as well as addition). ▲
Roman numerals
An effect to be reckoned with multiplication is that it can easily lead to very large numbers, and an old numbering system such as Roman numerals, create really big problems. For example
DXXXXVIII CCCLXVII x (548 x 367) gives
___
CCMCXVI , (201,116)
but to make this calculation I have to get help from some mechanical device: Here it is! ▲
The Abaco (or bullets)
The abacus, or abacus, was used to make the most of account, you can do addition and subtraction, multiplication, and even divisions: let's see 'how they could play with this medium that I have set out the multiplication above. The method is the same whether you are using decimal or Roman numerals: the payments are made directly on the abacus (quindi non occorre trascrivere i risultati intermedi, cosa che con i numeri romani sarebbe faticosissimo); alla fine del calcolo il risultato potrà essere direttamente trascritto in una qualunque delle due notazioni, decimale o romana.
Nelle figure che seguono faccio vedere i vari passaggi; a destra mostro sempre i fattori che si stanno moltiplicando, evidenziando in arancione le cifre coinvolte nel calcolo. A fianco delle righe del pallottoliere indico la somma da fare volta per volta sul pallottoliere stesso.
Quindi comincio con il pallottoliere "azzerato" (tutti i pallini a sinistra) e trascrivo il prodotto delle due cifre delle unità, 8 x 7 = 56.
As you see, now the bottom row (unit) is the value 6 (dots move to the right) and the second line (tens) is the value 5: then 5 tens plus 6 units is 56 .
Now I have to add 4 x 7 = 28, but above a line (like you do the multiplication in the column, moving to the left). The problem is that there is to be added in the second row 5 with 8, which gives carryover here is that in the second put a 3 (the units of the number 13, see diagram below left), move a ball in the third row indicates carry (central pattern), and finally the second highest in the third row: result of this first sum is 336.
The addition of 35 to the third and fourth line is not a problem (there are no carry-over):
followed in sequence all the missing steps:
transcribing the result we have two hundred thousand, zero tens of thousands, a thousand, one hundred, ten and six units. In other "words" just
___
CCMCXVI (201,116)
Systems like this were the only possible having to do multiplication with Roman numerals. But still there are people who are faster on the abacus to do multiplications and on the calculator: it is obviously a matter of habit! ▲
The method of "Prostaferesi"
The system of abacus was definitely good for the merchants and money changers, but it was far too complicated and dispersed to scientists. Dispersive because when multiplying decimal numbers, the number of digits after the decimal point increases enormously: for example, if I multiply two numbers with respectively 4 and 5 decimal places, the exact result it includes 9; scientists such as Kepler, who worked not on numbers but on accurate measurements made "eye" of the movement of planets in the sky, restricted their calculations to a reasonable number of decimal places, then use systems that can calculate them all, but really all these places (as in the case of the abacus) was really a waste of time.
In the sixteenth century astronomers felt more than ever the need to perform calculations faster, even if loosely, and this for various reasons: first trips to America, recently discovered, called for ocean routes to be followed much more complex than the simple cabotage along the coasts, and the Navy asked the scientists a safe (and possibly easier) to determine the point ship. Also in 1543 Nicolaus Copernicus published his famous book "De revolutionibus orbium coelestium" which assumed a heliocentric solar system type, some scientists (including Kepler, read below) felt led to study the heavens again to understand the real operation.
All these things require multiplication, and going to finish that scientists spent the bulk of the their time doing "simple" multiplication! Hence the need to invent new methods.
The first attempt to simplify the multiplication was done by triangulation: around 1580 was devised an ingenious system, called "prostaferesi", which helps you convert a multiplication in a series of simple steps, the procedure was based on a formula found by the German mathematician Johann Werner
cos (α) x cos (β) = ½ [cos (α + β) + cos (α - β)]
In this formula we see appear to four times the symbol " cos ", which stands for "Cosine". Here we do not need to know exactly what it is: we just know that since the time of Ptolemy in which the tables were compiled for each angle were calculated, once and for all, the various trigonometric functions fundamental enough to consult them convert an angle in his or cosine, given a numeric value, find the angle whose cosine was just the numeric value given.
better insight into the formula which I quoted above: to the left of the equal appears the product of cosine functions of two angles α and β, on the right shows the sum of cosine functions between the sum and difference of these angles. Here's how this formula was used to calculate the multiplication: Let's say we want to compute
0.91852 x 0.96963 (the exact result is 0.8906245476). If I say that
cos (α) = 0.91852 and
cos (β) = 0.96963
my formula becomes:
0.91852 x 0.96963 = ½ [cos (α + β) + cos (α - β)]
The equal sign tells me that the expression on his left (the product I want to calculate) that is exactly what the expression on his right, but to evaluate the expression on the right I need α and β values: to get just consult the appropriate tables (I reproduce the following images from the book of tables numeriche che usavo alle superiori)
Il valore di α è l'arco il cui coseno vale 0,91852. Cerco nella tabella a sinistra quale numero nella colonna del coseno si avvicina di più a questo valore, e trovo un angolo di 23° 17'. Analogamente faccio per β, il cui coseno vale 0,96963, quindi trovo l'angolo 14° 9'.
Ottenuti i valori degli angoli α e β posso calcolare la loro somma e differenza:
α + β = 23° 17' + 14° 9' = 37° 26'
α - β = 23° 17' - 14° 9' = 9° 8 '
Of these new values \u200b\u200bdo I need cosines, then go back to my table:
Here then
cos (α + β) = cos (23 ° 17' + 14 ° 9 ') = cos (37 ° 26') = 0.79406
cos (α - β) = cos (23 ° 17 '- 14 ° 9') = cos (09 ° 08 ') = 0.98732
The formula asks me to calculate the average of these two values, that is half their sum. We do the calculation:
0.91856 x 0.96966 = ½ [cos (α + β) + cos (α - β ] ½ = [0.79406 + 0.98732] = 0.89069.
Compared to the true value, which is 0.8906245476, I get an error of approximately 0.000065: very good approximation!
summary: given two values \u200b\u200bto multiply, find the angles of which are cosine; high and subtract, find the cosine of these two new values, the half-sum calculation: this is the product sought. Obviously, the method becomes more complicated if the factors are greater than one, in which case, a few more steps, but despite this, astronomers of the sixteenth century, much preferred to use that system rather than compute the multiplication on the card or the abacus! ▲
I "rods Napier"
method prostaferesi will soon be replaced by the use of "logarithms" of which I speak here. The same author of the latter, John Napier also invented an ingenious system of calculation, can do as well as multiplication and division, even square roots.
It starts with two rulers of wood mounted at 90 degrees as shown below:
and a series of poles (here we show 9):
Each stick is identified by a figure at the top, da 1 a 9. Di sotto ciascuno dei bastoncini ha nove caselle, che riportano il prodotto della cifra in alto con i numeri che vanno da 1 a 9; le cifre delle decine sono separate dalle cifre delle unità da una riga diagonale.
Per vedere come funziona il tutto, la cosa più veloce è fare un esempio pratico. Dovendo calcolare 845 x 6 (= 5070) basta disporre i bastoncini così:
Ora va considerata la sola riga che ha il numero 6 sulla sinistra, e osservato il contenuto delle caselle di ciascun bastoncino partendo da destra.
Nell'immagine the left, the figure highlighted in yellow represents the unity of the product 6 x 5 = 30, which will be the units digit of the final calculation. On the right we see highlighted the tens of the same product, together with the units digit of the product 6 x 4 = 24: the sum of 3 and 4 gives 7, which is the tens of final product.
Go ahead:
Now we need to add the numbers 2 and 8, giving 10: the hundreds digit of the result is 0, but I must remember that there is to take into account a carry. On the right we have only a 4: The high carry-over that I had open and I get the thousands digit of the result. Total calculated: 5070.
Let us now as a bit 'more complex: let's try 836 x 874 523. As we will see, you should write down the intermediate steps, but let's start by placing the bars:
Now we proceed to increase the figure to 874,523 units 836:
The result, occurs easily, 5247138. Then comes the turn of the figure 3:
Result 2623569. Finally, the figure 8:
result 6996184. Now we do the sum (hand):
+
2623569 5247138 6996184
----------------
731101228
which is the exact result! ▲
Next Chapter: Division
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