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If now we compute the anti-logs. of .00001, .00002, &c. sucessively, we shall have constantly D= log. m log. n=.00001, and the formula will be

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where m and n are any two numbers whose logarithms differ by

.00001.

Let log.m=.00001 and log. n=0, whence n=1, then m or the anti-logarithm of .00001 is found by (80) to be

A. Log..00001= 1.00001,00000,5.

Let log.m=.00002, log. n=.00001; then n=A. Log..00001= 1.00001,00000,5; and (80) gives

=

A. Log..00002 (1.00001,00000,5).

We shall find in the same manner

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The construction of the table is thus reduced to finding the succes

sive

powers of the number 1.00001,00000,5, an operation of little difficulty. The calculation would be as follows:

A. Log..00001=1.00001,00000,5

1,00001,0

5

A. Log..00002-1.00002,00002,0

1,00002,0

5

A. Log..00003-1.00003,00004,5

1,00003,0

5

A. Log..00004-1.00004,00008,0

As this operation is carried on entirely by addition, the accumulated error arising from the neglected terms of the series may at length affect the last place of decimals retained. It will be proper, therefore, to calculate separately the anti-logs. of .00100, .00200, .00300, &c., which may be done by the method employed above. These will serve as test or proof numbers; and the intervening antilogs. will be found by renewing the computation at .00100, .00200, &c. The error arising from the neglected figures will thus be corrected, and will not accumulate through more than 100 terms.

123. If we apply the formula to the common system, we make M=.434, &c., and by taking constantly D=.00001, as before, we D shall have .00002,30258,509. The formula will become

M

m=nx1.00002,30261,16,

and we shall find as above

A. Log..00001=1.00002,30261,16,

A. Log..00002 (1.00002,30261,16)3,

A. Log..00003 (1.00002,30261,16)3,

&c.

&c.

For other methods of constructing anti-logarithms, the student is referred to Hutton's History of Logarithms, prefixed to the former editions of his "Mathematical Tables."

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