1. Rather large number - oops
- Posted by Molasses <molasses at ALPHALINK.COM.AU> Nov 25, 1998
- 396 views
------=_NextPart_000_0012_01BE1876.00316FC0 charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable Thanks to those that answered, but I think the equation should have read : (16777216^480000) / 756864000 16m to the power of 480000, instead of the other way round, sorry. Of course (not that I understood any of the math of it), as Hawke said, = 3.63e14 would be a better divisor. The equation would then be: 256^3^480000/3.63e14 is this right? and could someone answer it please? TIA -molasses ------=_NextPart_000_0012_01BE1876.00316FC0 charset="iso-8859-1" Content-Transfer-Encoding: quoted-printable <!DOCTYPE HTML PUBLIC "-//W3C//DTD W3 HTML//EN"> <HTML> <HEAD> <META content=3Dtext/html;charset=3Diso-8859-1 = http-equiv=3DContent-Type> <META content=3D'"MSHTML 4.72.2106.6"' name=3DGENERATOR> </HEAD> <BODY bgColor=3D#ffffff> <DIV><FONT size=3D2>Thanks to those that answered,</FONT></DIV> <DIV><FONT color=3D#000000 size=3D2>but I think the equation should have = read=20 :</FONT></DIV> <DIV><FONT color=3D#000000 size=3D2></FONT> </DIV> <DIV><FONT size=3D2>(16777216^480000) / 756864000</FONT></DIV> <DIV><FONT size=3D2></FONT> </DIV> <DIV><FONT size=3D2>16m to the power of 480000, instead of the other way = round,=20 sorry.</FONT></DIV> <DIV><FONT color=3D#000000 size=3D2>Of course (not that I understood any = of the math=20 of it), as Hawke said, 3.63e14 would be a better divisor.</FONT></DIV> <DIV> </DIV> <DIV><FONT size=3D2>The equation would then be:</FONT></DIV> <DIV>256^3^480000/3.63e14</DIV> <DIV> </DIV> <DIV>is this right?</DIV> <DIV><FONT color=3D#000000 size=3D2>and could someone answer it please?=20 </FONT></DIV> <DIV> </DIV> <DIV><FONT color=3D#000000 size=3D2>TIA</FONT></DIV> <DIV><FONT color=3D#000000 size=3D2></FONT> </DIV> ------=_NextPart_000_0012_01BE1876.00316FC0--
2. Re: Rather large number - oops
- Posted by Rolf Schroeder <schroeder at DESY.DE> Nov 25, 1998
- 426 views
Molasses wrote: > > Thanks to those that answered, > but I think the equation should have read : > > (16777216^480000) / 756864000 > > 16m to the power of 480000, instead of the other way round, sorry. > Of course (not that I understood any of the math of it), as Hawke > said, 3.63e14 would be a better divisor. > > The equation would then be: > 256^3^480000/3.63e14 > > is this right? > and could someone answer it please? > > TIA > > -molasses OK, again: (look at the end!) -- -- -- d = (16777216)^480000 / 3.63e14 -- -- using log: log(d) = 480000 * log(16777216) - log(3.63e14) -- dec. log: dlog(d) = log(d)/log(10) -- atom a, b, c atom ln, ld, dec_exponent, dlog_mantissa, mantissa a = 480000 b = 16777216 c = 756864000 ln = b * log(a) - log(c) ld = ln/log(10) printf(1,"Dec. log : %25.15f\n", ld) dec_exponent = floor(ld) printf(1,"Dec.exp. : %25.15f\n", dec_exponent) dlog_mantissa = ld - dec_exponent printf(1,"DLog.Mantissa: %25.15f\n", dlog_mantissa) mantissa = power(10,dlog_mantissa) printf(1,"Mantissa : %25.15f\n", mantissa) -- -- The result of your number d is: -- puts(1,"The result:\n\n") printf(1,"(16777216)^480000 / 3.63e14 = %10.5f*10^%d\n", {mantissa,dec_exponent}) -- -- -- This will be the result: -- -- (16777216)^480000 / 3.63e14 = 3.22507*10^95315402 -- Have a nice day, Rolf
3. Re: Rather large number - oops
- Posted by Rolf Schroeder <schroeder at DESY.DE> Nov 25, 1998
- 522 views
Molasses wrote: > > Thanks to those that answered, > but I think the equation should have read : > > (16777216^480000) / 756864000 > > 16m to the power of 480000, instead of the other way round, sorry. > Of course (not that I understood any of the math of it), as Hawke > said, 3.63e14 would be a better divisor. > > The equation would then be: > 256^3^480000/3.63e14 > > is this right? > and could someone answer it please? > > TIA > > -molasses Excuse me mixed up the numbers! Here my second calculation: -- -- -- d = (16777216)^480000 / 3.63e14 -- -- using log: log(d) = 480000 * log(16777216) - log(3.63e14) -- dec. log: dlog(d) = log(d)/log(10) -- atom a, b, c atom ln, ld, dec_exponent, dlog_mantissa, mantissa a = 16777216 b = 480000 c = 3.63e14 ln = b * log(a) - log(c) ld = ln/log(10) printf(1,"Dec. log : %25.15f\n", ld) dec_exponent = floor(ld) printf(1,"Dec.exp. : %25.15f\n", dec_exponent) dlog_mantissa = ld - dec_exponent printf(1,"DLog.Mantissa: %25.15f\n", dlog_mantissa) mantissa = power(10,dlog_mantissa) printf(1,"Mantissa : %25.15f\n", mantissa) -- -- The result of your number d is: -- puts(1,"The result:\n\n") printf(1,"(16777216)^480000 / 3.63e14 = %10.5f*10^%d\n", {mantissa,dec_exponent}) -- -- -- This will be the result: -- -- (16777216)^480000 / 3.63e14 = 9.77558*10^3467850 -- Have a nice day, Rolf