Difference between revisions of "User:Regina/MyDrafts"

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(Converting between Number Systems)
(sensitivity of trigonometric functions)
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== Sensitivity ==
 
== Sensitivity ==
  
''Stabilität des Algorihtmus''
+
''Stabilität des Algorithmus''
 +
Todo
  
''Wie groß sind die Fehler, wenn die Eingangswerte nicht exakt sind?''
+
'''Hidden Bits'''
  
 
How much changes the result, if the input varies with one, two, three... bit in the internal representation?
 
How much changes the result, if the input varies with one, two, three... bit in the internal representation?
 +
 +
Trigonometric functions are very sensitive for huge input values.
 +
 +
Example of formula imput
 +
{| class="wikitable"
 +
|-
 +
! !! A !! B !! C
 +
|-
 +
! 1
 +
| =2^44-0.004 || =sin(A1) ||
 +
|-
 +
! 2
 +
| =2^44 || =sin(A2) || 
 +
|-
 +
! 3
 +
| =2^44+0.004 || =sin(A3) ||
 +
|-
 +
! 4
 +
|  ||  ||
 +
|}
 +
 +
Results shown in 15 figures precision
 +
{| class="wikitable"
 +
|-
 +
! !! A !! B !! C
 +
|-
 +
! 1
 +
| 17592186044416.00
 +
|| 0.386569623644289
 +
||
 +
|-
 +
! 2
 +
| 17592186044416.00
 +
|| 0.390169244205272
 +
|| 
 +
|-
 +
! 3
 +
| 17592186044416.00
 +
|| 0.393762911263629
 +
||
 +
|-
 +
! 4
 +
|  ||  ||
 +
|}
 +
 +
The correct value rounded to 15 figures precision would be sin(2^44) = 0.390169223351877
 +
 +
The trigonometric functions are reliable for arguments up to 2^27 with at least 13 figures accuracy. For larger values the accuracy decreases to about 6 figures accuracy for integral arguments and 2 figures accuracy for fractional values reaching 2^44. Huge arguments in the area above 2^44 might be totally wrong. If you need trigonometric functions for huge arguments, a spreadsheet is the wrong application for you.
 +
 +
The trigonometric functions are similar sensitive for arguments near their roots, but for those arguments the absolute result is near zero. So you will not notice the error, if you work with rounding to 15 or less decimal places.

Revision as of 19:47, 18 May 2010

Accuracy

This article is about situations, where you think, that Calc calculates wrong. You might have found a bug, and then you should write an issue. But look here first; perhaps the unexpected results come from something, you haven't been aware.

Precision in Calc

Calc uses for its calculation floating point numbers in double precision as defined in IEEE 754 standard. You get the best representation in a spreadsheet cell using the scientific format with format code 0.00000000000000E+000.

Calc rounds to two decimals in the default settings.

Beispieltabelle

Hinweis auf Berechnen wie angezeigt


But although you can force Calc to show 15 decimal digits, these might not be all accurate. The following sections list some of the problems.

Converting between Number Systems

Because a binary format is used internally, the numbers in internal calculation might differ slightly from the shown decimal values.

Most non integer numbers have infinite decimal places in binary format, which have to be rounded somewhere. Calculating with rounded values and converting back to decimal format gives different values then calculating manually in decimal format.

A1=0,2 A2=-5 A3=A2+$A$1 runterziehen. Ab Zelle A22 ist der Fehler sichtbar.

Tritt auch bei A1=-5 und A2=A1+0,2 auf

Alternative A2=$A$1+((ROW()-ROW(§A$1))*2)/10. Trick: möglichst mit ganzen Zahlen rechnen und erst zum Schluss in Dezimalzahlen umrechnen.

Known problem: If you fill a series based on two decimal numbers by dragging the bottom right handle of the marked cells, the generated values are not as accurate as they must be, see Issue 88119 . You get a better series by generating the values via Edit > Fill > Series, but still some values are not accurate in the last digit. You should control the results and correct them manually.

No Symbolic π

From mathematics you know sin(π)= 0 and you know that tan(π/2) is undefined. But you cannot get these results in Calc, because the value π is always treated as rounded floating point number. It makes no difference using PI() or RADIANDS(180). Calc cannot evaluate π symbolically as computer algebra systems do. That is no special limitation of Calc, but other often used spreadsheet applications work only numerically too.

A B C
1 1.63317787283838E+016 =TAN(PI()/2)
2 1.22460635382238E-016 =SIN(RADIANS(180))
3

No Fractional Arithmetic

Calc kann zwar Zahlen als Brüche darstellen, aber nicht mit ihnen rechnen.

Cancellation

If you subtract two non integer numbers, which have nearly the same value, the result has less significant digits then the initial values.

A B C
1 9.99411764795882E-001 =0.999411764795882
2 9.99411764705882E-001 =1699/1700
3 8.99997854020285E-011 =A1-A2
4 8.99996470588235E-011

Cell A4 shows the correct result of , calculated with a computer algebra system with high precision.

A B C
1 0.99999876543210000000
2 0.00000123456790002141 =1-A1
3
4

Calculating manually gives 1−0.99999876543210000000 = 0.00000123456790000000


Ill-conditioned problems

Verhalten bei Polstellen

Sensitivity

Stabilität des Algorithmus Todo

Hidden Bits

How much changes the result, if the input varies with one, two, three... bit in the internal representation?

Trigonometric functions are very sensitive for huge input values.

Example of formula imput

A B C
1 =2^44-0.004 =sin(A1)
2 =2^44 =sin(A2)
3 =2^44+0.004 =sin(A3)
4

Results shown in 15 figures precision

A B C
1 17592186044416.00 0.386569623644289
2 17592186044416.00 0.390169244205272
3 17592186044416.00 0.393762911263629
4

The correct value rounded to 15 figures precision would be sin(2^44) = 0.390169223351877

The trigonometric functions are reliable for arguments up to 2^27 with at least 13 figures accuracy. For larger values the accuracy decreases to about 6 figures accuracy for integral arguments and 2 figures accuracy for fractional values reaching 2^44. Huge arguments in the area above 2^44 might be totally wrong. If you need trigonometric functions for huge arguments, a spreadsheet is the wrong application for you.

The trigonometric functions are similar sensitive for arguments near their roots, but for those arguments the absolute result is near zero. So you will not notice the error, if you work with rounding to 15 or less decimal places.

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