APL(NARS), 92 chars
r←a phi b;e
r←b⋄e←÷10v*⌊⎕FPC÷3.322
b←r⋄a←2÷⍨a+b⋄r←√a×ばつ⍳e<∣r-b
ln←{((2÷⍨⍵+1)phi√⍵)÷⍨⍵-1}
//12+23+31+26= 92
Here I would userewrite the function phi that one can find in the book
ESERCIZI E COMPLEMENTI DI ANALISI MATEMATICA
VOLUME PRIMO - AUTORE: ENRICO GIUSTI
in the end of "capitolo 3". The same function the Autor says is useful for calculate the function arccos() too.
The problem that I see, is the wrong build of float numeric in computer as the number is in memory the sum of decimal part and fractional part and the dimension in memory of decimal part influence the fractional part and so the precision...
I remember that there is one model in that make float point decimal part, not influence the fractional part as memory space... Is it that the fixed point model?
P is a function for print the number in number of digits that have to be ok if number
and its calculus not use numbers with decimal digits > ⎕fpc÷4 in my opinion... (it is as
⎕fpc÷4 is for decimal part and ⎕fpc÷4 is for fractional part , if the memory of
decimal part is too much the precision decrease). The numeber that end with the v
is the big float.
P←{⍵⍕⍨⌊⎕fpc÷8}
ln 0.1
̄2.302585093
P ln 0.1
̄2.3025850929940459
P ln 2
0.6931471805599454
P ln 99
4.5951198501345889
⎕fpc←256
P ln 99
4.5951198501345889________________
P ln 99v
4.59511985013458992685243405181018
⎕fpc←1024
P ln 99v
4.59511985013458992685243405181018070911668796958291607868795637640556221035045464682228621763009896899065578402005373000699157761
APL(NARS), 92 chars
r←a phi b;e
r←b⋄e←÷10v*⌊⎕FPC÷3.322
b←r⋄a←2÷⍨a+b⋄r←√a×ばつ⍳e<∣r-b
ln←{((2÷⍨⍵+1)phi√⍵)÷⍨⍵-1}
//12+23+31+26= 92
Here I would use the function phi that one can find in the book
ESERCIZI E COMPLEMENTI DI ANALISI MATEMATICA
VOLUME PRIMO - AUTORE: ENRICO GIUSTI
in the end of "capitolo 3". The same function the Autor says is useful for calculate the function arccos() too.
The problem that I see, is the wrong build of float numeric in computer as the number is in memory the sum of decimal part and fractional part and the dimension in memory of decimal part influence the fractional part and so the precision...
I remember that there is one model in that make float point decimal part, not influence the fractional part as memory space... Is it that the fixed point model?
P is a function for print the number in number of digits that have to be ok if number
and its calculus not use numbers with decimal digits > ⎕fpc÷4 in my opinion... (it is as
⎕fpc÷4 is for decimal part and ⎕fpc÷4 is for fractional part , if the memory of
decimal part is too much the precision decrease). The numeber that end with the v
is the big float.
P←{⍵⍕⍨⌊⎕fpc÷8}
ln 0.1
̄2.302585093
P ln 0.1
̄2.3025850929940459
P ln 2
0.6931471805599454
P ln 99
4.5951198501345889
⎕fpc←256
P ln 99
4.5951198501345889________________
P ln 99v
4.59511985013458992685243405181018
⎕fpc←1024
P ln 99v
4.59511985013458992685243405181018070911668796958291607868795637640556221035045464682228621763009896899065578402005373000699157761
APL(NARS), 92 chars
r←a phi b;e
r←b⋄e←÷10v*⌊⎕FPC÷3.322
b←r⋄a←2÷⍨a+b⋄r←√a×ばつ⍳e<∣r-b
ln←{((2÷⍨⍵+1)phi√⍵)÷⍨⍵-1}
//12+23+31+26= 92
Here I would rewrite the function phi that one can find in the book
ESERCIZI E COMPLEMENTI DI ANALISI MATEMATICA
VOLUME PRIMO - AUTORE: ENRICO GIUSTI
in the end of "capitolo 3". The same function the Autor says is useful for calculate the function arccos() too.
The problem that I see, is the wrong build of float numeric in computer as the number is in memory the sum of decimal part and fractional part and the dimension in memory of decimal part influence the fractional part and so the precision...
I remember that there is one model in that make float point decimal part, not influence the fractional part as memory space... Is it that the fixed point model?
P is a function for print the number in number of digits that have to be ok if number
and its calculus not use numbers with decimal digits > ⎕fpc÷4 in my opinion... (it is as
⎕fpc÷4 is for decimal part and ⎕fpc÷4 is for fractional part , if the memory of
decimal part is too much the precision decrease). The numeber that end with the v
is the big float.
P←{⍵⍕⍨⌊⎕fpc÷8}
ln 0.1
̄2.302585093
P ln 0.1
̄2.3025850929940459
P ln 2
0.6931471805599454
P ln 99
4.5951198501345889
⎕fpc←256
P ln 99
4.5951198501345889________________
P ln 99v
4.59511985013458992685243405181018
⎕fpc←1024
P ln 99v
4.59511985013458992685243405181018070911668796958291607868795637640556221035045464682228621763009896899065578402005373000699157761
APL(NARS), 92 chars
r←a phi b;e
r←b⋄e←÷10v*⌊⎕FPC÷3.322
b←r⋄a←2÷⍨a+b⋄r←√a×ばつ⍳e<∣r-b
ln←{((2÷⍨⍵+1)phi√⍵)÷⍨⍵-1}
//12+23+31+26= 92
Here I would use the function phi that one can find in the book
ESERCIZI E COMPLEMENTI DI ANALISI MATEMATICA
VOLUME PRIMO - AUTORE: ENRICO GIUSTI
in the end of "capitolo 3". The same function the Autor says is useful for calculate the function arccos() too.
The problem that I see, is the wrong build of float numeric in computer as the number is in memory the sum of decimal part and fractional part and the dimension in memory of decimal part influence the fractional part and so the precision...
I remember that there is one model in that make float point decimal part, not influence the fractional part as memory space... Is it that the fixed point model?
P is a function for print the number in number of digits that have to be ok if number
and its calculus not use numbers with decimal digits > ⎕fpc÷4 in my opinion... (it is as
⎕fpc÷4 is for decimal part and ⎕fpc÷4 is for fractional part , if the memory of
decimal part is too much the precision decrease). The numeber that end with the v
is the big float.
P←{⍵⍕⍨⌊⎕fpc÷8}
ln 0.1
̄2.302585093
P ln 0.1
̄2.3025850929940459
P ln 2
0.6931471805599454
P ln 99
4.5951198501345889
⎕fpc←256
P ln 99
4.5951198501345889________________
P ln 99v
4.59511985013458992685243405181018
⎕fpc←1024
P ln 99v
4.59511985013458992685243405181018070911668796958291607868795637640556221035045464682228621763009896899065578402005373000699157761