Skip to main content

Why does N not upgrade precision? 



Precision[N[1.0, 20]]
Precision[N[1, 20]]



MachinePrecision
20.

It would be so much more intuitive and less error prone, if Precision[N[1.0, 20]] would be 20 and not MachinePrecision. Why do I have to explicitly use N[Rationalized[1.0], 20] to upgrade the precision?


If it is about the warning messages during the calculations, Mathematica could return a warning already at the Precision[N[1.0, 20]] step.


Edit I am not attempting to use N for rounding. I also understand the difference between MachinePrecision and arbitrary precision, which is very well described in this answer.


I want to know, why does Mathematica not consider the description of the number existing in the underling binary representation as exact if I apply N, and why I need to wrap it with Rationalize.


Is it performance?



Is it some deep semantic meaning of N vs SetPrecision?


Is set SetPrecision any different from N@Rationalize@?



Answer



N can only lower precision, it cannot raise precision.


N[1.3`4, 10] //Precision


4.



Since MachinePrecision is considered to be the lowest possible precision for a number, applying N to a machine precision number does nothing.



Addendum


(Hopefully the following will resolve some of your confusion)


Let's start with your title. N does not upgrade precision because that would mean that N is making up digits so that the precision could increase. Making up digits to increase the precision of a number is a job for SetPrecision.


Think of N as taking an input with some validated amount of precision, and returning an output with a smaller validated amount of precision (approximating). By validated amount of precision, I'm referring to the precision tracking that goes on for arbitrary precision numbers. This is the value returned by Precision. N will never return a number with a larger validated amount of precision, because in order to do so, it would have to make up digits. So:


Precision[N[3`4, 10]]


4.



If N returned a number with 10 digits of precision, those digits would be completely made up.



Therefore, you can trust the value returned by N. If the value returned by N had 10 digits of precision, than the input had enough precision so that a 10 digit approximation could be obtained. If the value returned by N had less than 10 digits of precision, than the input did not have enough precision to determine a 10 digit approximation.


Finally, let's consider machine numbers. It should be clear that machine numbers have no precision tracking, so the validated amount of precision is 0. There is no way for N to convert such numbers into arbitrary precision numbers with some validated amount of precision.


Comments

Post a Comment

Popular posts from this blog

plotting - How to draw lines between specified dots on ListPlot?

I would like to create a plot where I have unconnected dots and some connected. So far, I have figured out how to draw the dots. My code is the following: ListPlot[{{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4,13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full] I have thought using ListLinePlot command, but I don't know how to specify to the command to draw only selected lines between the dots. Do have any suggestions/hints on how to do that? Thank you. Answer One possibility would be to use Epilog with Line : ListPlot[ {{1, 1}, {2, 2}, {3, 3}, {4, 4}, {1, 4}, {2, 5}, {3, 6}, {4, 7}, {1, 7}, {2, 8}, {3, 9}, {4, 10}, {1, 10}, {2, 11}, {3, 12}, {4, 13}, {2.5, 7}}, Ticks -> {{1, 2, 3, 4}, None}, AxesStyle -> Thin, TicksStyle -> Directive[Black, Bold, 12], Mesh -> Full, Epilog -> { Line[ ...
Post a Comment
Read more

equation solving - Invert and fit implicitly defined curve

I need to fit an implicitly defined curve. I thought I could get some data out of Solve , and then using FindFit . Therefore, I would like to find the relation the parametric curve defined by $F(x,y)=0$: Solve[-(1/2) + 1/2 (0.41202 BesselK[0, 0.1 Sqrt[x^2 + y^2]] + (0.101483 x BesselK[1, 0.1 Sqrt[x^2 + y^2]])/Sqrt[x^2 + y^2]) == 0, y] But I can't get an output: Solve was unable to solve the system with inexact coefficients or the system obtained by direct rationalization of inexact numbers present in the system. Since many of the methods used by Solve require exact input, providing Solve with an exact version of the system may help. >> Edit: In particular, I would like to fit the data coming from the curve with the expression of another curve, and not with a function $f(x)$. In particular, since this clearly looks like a cardioid , I would like it to fit to something like it. What other strategies could I try?
Post a Comment
Read more

dynamic - How can I make a clickable ArrayPlot that returns input?

I would like to create a dynamic ArrayPlot so that the rectangles, when clicked, provide the input. Can I use ArrayPlot for this? Or is there something else I should have to use? Answer ArrayPlot is much more than just a simple array like Grid : it represents a ranged 2D dataset, and its visualization can be finetuned by options like DataReversed and DataRange . These features make it quite complicated to reproduce the same layout and order with Grid . Here I offer AnnotatedArrayPlot which comes in handy when your dataset is more than just a flat 2D array. The dynamic interface allows highlighting individual cells and possibly interacting with them. AnnotatedArrayPlot works the same way as ArrayPlot and accepts the same options plus Enabled , HighlightCoordinates , HighlightStyle and HighlightElementFunction . data = {{Missing["HasSomeMoreData"], GrayLevel[ 1], {RGBColor[0, 1, 1], RGBColor[0, 0, 1], GrayLevel[1]}, RGBColor[0, 1, 0]}, {GrayLevel[0], GrayLevel...
Post a Comment
Read more