Bošković, Ruđer Josip
,
Abhandlung von den verbesserten dioptrischen Fernröhren aus den Sammlungen des Instituts zu Bologna sammt einem Anhange des Uebersetzers
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Anhang
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Reihe. </
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<
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xml:space
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">1 multiplicirt mit {1/m} (welches man
<
lb
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unter a findet) giebt {1/m}: </
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<
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xml:id
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echoid-s2017
"
xml:space
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preserve
">dieß ſchreibe man
<
lb
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für das dritte Glied der dritten Reihe. </
s
>
<
s
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echoid-s2018
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xml:space
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">Fer-
<
lb
/>
ner wird das vierte {1/m} X - 1, das fünfte
<
lb
/>
- {1/m} X m = - 1, das ſechſte - 1 X -
<
lb
/>
1 = 1, das ſ@ebente 1 X {1/m}, das achte
<
lb
/>
{1/m} X m = 1. </
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>
<
s
xml:id
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echoid-s2019
"
xml:space
="
preserve
">Man hat alſo folgende drey
<
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Reihen:
<
lb
/>
</
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<
s
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echoid-s2020
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xml:space
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">d : </
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<
s
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echoid-s2021
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xml:space
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">c: </
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<
s
xml:id
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echoid-s2022
"
xml:space
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">b: </
s
>
<
s
xml:id
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echoid-s2023
"
xml:space
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">a: </
s
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<
s
xml:id
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echoid-s2024
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xml:space
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">b: </
s
>
<
s
xml:id
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echoid-s2025
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xml:space
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">c: </
s
>
<
s
xml:id
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echoid-s2026
"
xml:space
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">b: </
s
>
<
s
xml:id
="
echoid-s2027
"
xml:space
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">a
<
lb
/>
{1/M}:</
s
>
<
s
xml:id
="
echoid-s2028
"
xml:space
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">M:</
s
>
<
s
xml:id
="
echoid-s2029
"
xml:space
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">{1/m}: </
s
>
<
s
xml:id
="
echoid-s2030
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xml:space
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">- 1: </
s
>
<
s
xml:id
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echoid-s2031
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xml:space
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">m: </
s
>
<
s
xml:id
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echoid-s2032
"
xml:space
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preserve
">- 1: </
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>
<
s
xml:id
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echoid-s2033
"
xml:space
="
preserve
">{1/m}: </
s
>
<
s
xml:id
="
echoid-s2034
"
xml:space
="
preserve
">m
<
lb
/>
{1/M}: </
s
>
<
s
xml:id
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echoid-s2035
"
xml:space
="
preserve
">1: </
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>
<
s
xml:id
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echoid-s2036
"
xml:space
="
preserve
">{1/m}: </
s
>
<
s
xml:id
="
echoid-s2037
"
xml:space
="
preserve
">- {1/m}: </
s
>
<
s
xml:id
="
echoid-s2038
"
xml:space
="
preserve
">- 1: </
s
>
<
s
xml:id
="
echoid-s2039
"
xml:space
="
preserve
">1: </
s
>
<
s
xml:id
="
echoid-s2040
"
xml:space
="
preserve
">{1/m}: </
s
>
<
s
xml:id
="
echoid-s2041
"
xml:space
="
preserve
">1</
s
>
</
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<
p
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<
s
xml:id
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echoid-s2042
"
xml:space
="
preserve
">Nun ziehe man 1 von dem erſten Gliede
<
lb
/>
der zweyten Reihe {1/M} ab, ſo wird {1/M} - 1
<
lb
/>
= {1-M/M} = {M-1/-M}; </
s
>
<
s
xml:id
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echoid-s2043
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xml:space
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preserve
">dieſes dividire man
<
lb
/>
mit d X {1/M} = {d/M}, ſo bekommt man den
<
lb
/>
erſten Bruch {M-1/-M} X {M/d} = {M-1/M-d}.
<
lb
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</
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>
<
s
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xml:space
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preserve
">Nach dieſem zieht man von (dem zwey-
<
lb
/>
ten Gliede der zweyten Reihe) 1 ab, und
<
lb
/>
dividirt den Ueberſchuß mit c X 1: </
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xml:space
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