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人民教育出版社編譯的五位對數表中第一表腳下有一個附表。這個表的用法,書中已有說明,現在我把它的理由說一說。因故當α甚小而求Sinα的近似值時,可以α的弧度數代Sinα如欲求logSin1°9′46″,查原書第一表下的附表,1°9′46″與1°10′最近,即用下式。(Sin1°9′46″)/(Sin1°10′)=(π/(180×60~2)×(1°9′46″的秒數)/(π/(180×60~2)×(1°10′的秒數))=(1°9′46″的秒數)/(1°10′的秒數),因之 logSin1°9′46″=logSin1°10′-log(1°10′的秒數)+log(1°9′46″的秒數)。原書第一表下之附表中所標9下之數,系此式中右端前兩項的代數和再加以10。即如附表中列1°10′這一行的S下所記的數是4.68554,系照下麵求得的;
There is a schedule at the foot of the first table in the five-digit log compiled by People’s Education Press. The usage of this table has already been explained in the book. Now I will explain its reasons. For the sake of the fact that when α is small and the approximate value of Sinα is found, the radians of α can be substituted for Sinα. For example, if you want to find logSin1°9′46′′, check the schedules in the first table of the original book, 1°9′46′′ and 1°10′. Recently, the following formula is used. (Sin1°9′46′′)/(Sin1°10′)=(π/(180×60~2)×(1°9′46′′ seconds)/(π/(180×60~2)× (seconds in 1°10′))=(1°9′46′′ seconds)/(1°10′seconds), because of logSin1°9′46′′=logSin1°10′-log(1) °10’s seconds + log (seconds of 1°9′46′′.) The number in the attached table under the first table of the original book is the number of the first two items in the formula, which is the algebra and addition of the first two items in the right hand side of the formula. 10. That is, if the number recorded under S in the line of 1°10′ in the attached table is 4.68554, it is obtained as follows;