# Page:Scientific Memoirs, Vol. 3 (1843).djvu/696

But when ${\displaystyle \scriptstyle {n}}$ is given for the particular case to be calculated, it will be further requisite that the machine limit the number of its multiplications according to the given values. The process may be thus arranged. The three numbers ${\displaystyle \scriptstyle {a}}$, ${\displaystyle \scriptstyle {b}}$ and ${\displaystyle \scriptstyle {n}}$ will be written on as many distinct columns of the store; we shall designate them ${\displaystyle \scriptstyle {\mathbf {V} _{0}}}$, ${\displaystyle \scriptstyle {\mathbf {V} _{1}}}$, ${\displaystyle \scriptstyle {\mathbf {V} _{2}}}$; the result ${\displaystyle \scriptstyle {ab^{n}}}$ will place itself on the column ${\displaystyle \scriptstyle {\mathbf {V} _{3}}}$. When the number ${\displaystyle \scriptstyle {n}}$ has been introduced into the machine, a card will order a certain registering-apparatus to mark ${\displaystyle \scriptstyle {(n-1)}}$, and will at the same time execute the multiplication of ${\displaystyle \scriptstyle {b}}$ by ${\displaystyle \scriptstyle {b}}$. When this is completed, it will be found that the registering-apparatus has effaced a unit, and that it only marks ${\displaystyle \scriptstyle {(n-2)}}$; while the machine will now again order the number ${\displaystyle \scriptstyle {b}}$ written on the column ${\displaystyle \scriptstyle {\mathbf {V} _{1}}}$ to multiply itself with the product ${\displaystyle \scriptstyle {b^{2}}}$ written on the column ${\displaystyle \scriptstyle {\mathbf {V} _{3}}}$, which will give ${\displaystyle \scriptstyle {b^{3}}}$. Another unit is then effaced from the registering-apparatus, and the same processes are continually repeated until it only marks zero. Thus the number ${\displaystyle \scriptstyle {b^{n}}}$ will be found inscribed on ${\displaystyle \scriptstyle {\mathbf {V} _{3}}}$, when the machine, pursuing its course of operations, will order the product of ${\displaystyle \scriptstyle {b^{n}}}$ by ${\displaystyle \scriptstyle {a}}$; and the required calculation will have been completed without there being any necessity that the number of operation-cards used should vary with the value of ${\displaystyle \scriptstyle {n}}$. If ${\displaystyle \scriptstyle {n}}$ were negative, the cards, instead of ordering the multiplication of ${\displaystyle \scriptstyle {a}}$ by ${\displaystyle \scriptstyle {b^{n}}}$, would order its division; this we can easily conceive, since every number, being inscribed with its respective sign, is consequently capable of reacting on the nature of the operations to be executed. Finally, if ${\displaystyle \scriptstyle {n}}$ were fractional, of the form ${\displaystyle \scriptstyle {\frac {p}{q}}}$, an additional column would be used for the inscription of ${\displaystyle \scriptstyle {q}}$, and the machine would bring into action two sets of processes, one for raising ${\displaystyle \scriptstyle {b}}$ to the power ${\displaystyle \scriptstyle {p}}$, the other for extracting the ${\displaystyle \scriptstyle {q}}$th root of the number so obtained.
Again, it may be required, for example, to multiply an expression of the form ${\displaystyle \scriptstyle {ax^{m}+bx^{n}}}$ by another ${\displaystyle \scriptstyle {{\text{A}}x^{p}+{\text{B}}x^{q}}}$, and then to reduce the product to the least number of terms, if any of the indices are equal. The two factors being ordered with respect to ${\displaystyle \scriptstyle {x}}$, the general result of the multiplication would be ${\displaystyle \scriptstyle {{\text{A}}ax^{m+p}+{\text{A}}bx^{n+p}+{\text{B}}ax^{m+q}+{\text{B}}bx^{n+q}}}$. Up to this point the process presents no difficulties; but suppose that we have ${\displaystyle \scriptstyle {m=p}}$ and ${\displaystyle \scriptstyle {n=q}}$, and that we wish to reduce the two middle terms to