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Inconsistent results from Wolfram Could


Find Determinant/or Row Reduce parameter dependent matrixPack Solve results into a vectorHow to simplify symbolic matrix multiplication results?LUDecomposition does not give the expected resultsHow to interpret the results of PCAWhy do ReplaceAll and With give different results?Nearest non-collinear/non-coplanar pointsObtaining the determinant from a LinearSolveFunction objectRowReduction: Wolfram Alpha vs MathematicaEigenvectors calculation doesn't match from two identical results













1












$begingroup$


More specifically, I was using the "no sign-in" option of Wolfram Programming Lab



I was trying to solve a matrix problem, with the following code:



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=6.2832;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


Since the numerical values of w1 and w2 should be close, I expect the numerical values of D1 and D2 should also be close. Strangely, Wolfram Cloud gives very different values:



enter image description here



It took me a whole night to pin down this segment of code. I don't know if this is only due to my computer/browser, or some one else, if runs the same code, will have same problem? What happened?




Edit



Suppose I would like to compare the determinant using exact symbolic $2pi$ and function N[2Pi,5]



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=2Pi;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


The result is not exactly the same:
enter image description here



So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?










share|improve this question









New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$







  • 3




    $begingroup$
    Suppose small epsilon then ClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]] returns 12500000*epsilon*(epsilon - 2*Pi)*(epsilon + 4*Pi)*(epsilon + 6*Pi) and for small epsilon that is approximately 12500000*epsilon*-2*Pi*4*Pi*6*Pi== -600000000*epsilon*Pi^3` so any small error in w is multiplied by about 1.86*10^10 in the determinant.
    $endgroup$
    – Bill
    8 hours ago







  • 2




    $begingroup$
    No, N[x, p], represents, if possible, the value of x approximated to a precision of p digits. Read the documentation on N.
    $endgroup$
    – Michael E2
    2 hours ago






  • 2




    $begingroup$
    See reference.wolfram.com/language/tutorial/NumbersOverview.html, esp. the tutorials about exact, approximate and arbitrary-precision numbers.
    $endgroup$
    – Michael E2
    2 hours ago















1












$begingroup$


More specifically, I was using the "no sign-in" option of Wolfram Programming Lab



I was trying to solve a matrix problem, with the following code:



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=6.2832;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


Since the numerical values of w1 and w2 should be close, I expect the numerical values of D1 and D2 should also be close. Strangely, Wolfram Cloud gives very different values:



enter image description here



It took me a whole night to pin down this segment of code. I don't know if this is only due to my computer/browser, or some one else, if runs the same code, will have same problem? What happened?




Edit



Suppose I would like to compare the determinant using exact symbolic $2pi$ and function N[2Pi,5]



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=2Pi;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


The result is not exactly the same:
enter image description here



So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?










share|improve this question









New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$







  • 3




    $begingroup$
    Suppose small epsilon then ClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]] returns 12500000*epsilon*(epsilon - 2*Pi)*(epsilon + 4*Pi)*(epsilon + 6*Pi) and for small epsilon that is approximately 12500000*epsilon*-2*Pi*4*Pi*6*Pi== -600000000*epsilon*Pi^3` so any small error in w is multiplied by about 1.86*10^10 in the determinant.
    $endgroup$
    – Bill
    8 hours ago







  • 2




    $begingroup$
    No, N[x, p], represents, if possible, the value of x approximated to a precision of p digits. Read the documentation on N.
    $endgroup$
    – Michael E2
    2 hours ago






  • 2




    $begingroup$
    See reference.wolfram.com/language/tutorial/NumbersOverview.html, esp. the tutorials about exact, approximate and arbitrary-precision numbers.
    $endgroup$
    – Michael E2
    2 hours ago













1












1








1


1



$begingroup$


More specifically, I was using the "no sign-in" option of Wolfram Programming Lab



I was trying to solve a matrix problem, with the following code:



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=6.2832;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


Since the numerical values of w1 and w2 should be close, I expect the numerical values of D1 and D2 should also be close. Strangely, Wolfram Cloud gives very different values:



enter image description here



It took me a whole night to pin down this segment of code. I don't know if this is only due to my computer/browser, or some one else, if runs the same code, will have same problem? What happened?




Edit



Suppose I would like to compare the determinant using exact symbolic $2pi$ and function N[2Pi,5]



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=2Pi;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


The result is not exactly the same:
enter image description here



So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?










share|improve this question









New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.






$endgroup$




More specifically, I was using the "no sign-in" option of Wolfram Programming Lab



I was trying to solve a matrix problem, with the following code:



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=6.2832;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


Since the numerical values of w1 and w2 should be close, I expect the numerical values of D1 and D2 should also be close. Strangely, Wolfram Cloud gives very different values:



enter image description here



It took me a whole night to pin down this segment of code. I don't know if this is only due to my computer/browser, or some one else, if runs the same code, will have same problem? What happened?




Edit



Suppose I would like to compare the determinant using exact symbolic $2pi$ and function N[2Pi,5]



ClearAll["Global`*"]
m=2,0,0,1*2500;
k=3,-1,-1,1*20000 Pi^2;
w1=N[2Pi,5];
w2=2Pi;
D1=Det[k-w1^2*m]
D2=Det[k-w2^2*m]


The result is not exactly the same:
enter image description here



So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?







linear-algebra






share|improve this question









New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.










share|improve this question









New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








share|improve this question




share|improve this question








edited 2 hours ago







York Tsang













New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.








asked 8 hours ago









York TsangYork Tsang

1063




1063




New contributor



York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




New contributor




York Tsang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.









  • 3




    $begingroup$
    Suppose small epsilon then ClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]] returns 12500000*epsilon*(epsilon - 2*Pi)*(epsilon + 4*Pi)*(epsilon + 6*Pi) and for small epsilon that is approximately 12500000*epsilon*-2*Pi*4*Pi*6*Pi== -600000000*epsilon*Pi^3` so any small error in w is multiplied by about 1.86*10^10 in the determinant.
    $endgroup$
    – Bill
    8 hours ago







  • 2




    $begingroup$
    No, N[x, p], represents, if possible, the value of x approximated to a precision of p digits. Read the documentation on N.
    $endgroup$
    – Michael E2
    2 hours ago






  • 2




    $begingroup$
    See reference.wolfram.com/language/tutorial/NumbersOverview.html, esp. the tutorials about exact, approximate and arbitrary-precision numbers.
    $endgroup$
    – Michael E2
    2 hours ago












  • 3




    $begingroup$
    Suppose small epsilon then ClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]] returns 12500000*epsilon*(epsilon - 2*Pi)*(epsilon + 4*Pi)*(epsilon + 6*Pi) and for small epsilon that is approximately 12500000*epsilon*-2*Pi*4*Pi*6*Pi== -600000000*epsilon*Pi^3` so any small error in w is multiplied by about 1.86*10^10 in the determinant.
    $endgroup$
    – Bill
    8 hours ago







  • 2




    $begingroup$
    No, N[x, p], represents, if possible, the value of x approximated to a precision of p digits. Read the documentation on N.
    $endgroup$
    – Michael E2
    2 hours ago






  • 2




    $begingroup$
    See reference.wolfram.com/language/tutorial/NumbersOverview.html, esp. the tutorials about exact, approximate and arbitrary-precision numbers.
    $endgroup$
    – Michael E2
    2 hours ago







3




3




$begingroup$
Suppose small epsilon then ClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]] returns 12500000*epsilon*(epsilon - 2*Pi)*(epsilon + 4*Pi)*(epsilon + 6*Pi) and for small epsilon that is approximately 12500000*epsilon*-2*Pi*4*Pi*6*Pi== -600000000*epsilon*Pi^3` so any small error in w is multiplied by about 1.86*10^10 in the determinant.
$endgroup$
– Bill
8 hours ago





$begingroup$
Suppose small epsilon then ClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]] returns 12500000*epsilon*(epsilon - 2*Pi)*(epsilon + 4*Pi)*(epsilon + 6*Pi) and for small epsilon that is approximately 12500000*epsilon*-2*Pi*4*Pi*6*Pi== -600000000*epsilon*Pi^3` so any small error in w is multiplied by about 1.86*10^10 in the determinant.
$endgroup$
– Bill
8 hours ago





2




2




$begingroup$
No, N[x, p], represents, if possible, the value of x approximated to a precision of p digits. Read the documentation on N.
$endgroup$
– Michael E2
2 hours ago




$begingroup$
No, N[x, p], represents, if possible, the value of x approximated to a precision of p digits. Read the documentation on N.
$endgroup$
– Michael E2
2 hours ago




2




2




$begingroup$
See reference.wolfram.com/language/tutorial/NumbersOverview.html, esp. the tutorials about exact, approximate and arbitrary-precision numbers.
$endgroup$
– Michael E2
2 hours ago




$begingroup$
See reference.wolfram.com/language/tutorial/NumbersOverview.html, esp. the tutorials about exact, approximate and arbitrary-precision numbers.
$endgroup$
– Michael E2
2 hours ago










1 Answer
1






active

oldest

votes


















6












$begingroup$

I get the same result in Mathematica, so it's not a Mathematica Online issue. I don't think it's even a Mathematica issue. It's due to two factors:




  1. w1 is not equal to w2, because N doesn't actually truncate 2 Pi to five digits


  2. Det[k-w^2*m] changes quickly, so any little inaccuracy in w becomes a big discrepancy in Det[k-w^2*m]

To see #1:



w1 == 2 [Pi]
(* True *)
w1 - w2
(* -0.0000146928 *)


To see #2:



Plot[Det[k - w^2*m], w, 6.2831, 6.2833]


Mathematica graphics






share|improve this answer









$endgroup$












  • $begingroup$
    Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
    $endgroup$
    – York Tsang
    2 hours ago











Your Answer








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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









6












$begingroup$

I get the same result in Mathematica, so it's not a Mathematica Online issue. I don't think it's even a Mathematica issue. It's due to two factors:




  1. w1 is not equal to w2, because N doesn't actually truncate 2 Pi to five digits


  2. Det[k-w^2*m] changes quickly, so any little inaccuracy in w becomes a big discrepancy in Det[k-w^2*m]

To see #1:



w1 == 2 [Pi]
(* True *)
w1 - w2
(* -0.0000146928 *)


To see #2:



Plot[Det[k - w^2*m], w, 6.2831, 6.2833]


Mathematica graphics






share|improve this answer









$endgroup$












  • $begingroup$
    Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
    $endgroup$
    – York Tsang
    2 hours ago















6












$begingroup$

I get the same result in Mathematica, so it's not a Mathematica Online issue. I don't think it's even a Mathematica issue. It's due to two factors:




  1. w1 is not equal to w2, because N doesn't actually truncate 2 Pi to five digits


  2. Det[k-w^2*m] changes quickly, so any little inaccuracy in w becomes a big discrepancy in Det[k-w^2*m]

To see #1:



w1 == 2 [Pi]
(* True *)
w1 - w2
(* -0.0000146928 *)


To see #2:



Plot[Det[k - w^2*m], w, 6.2831, 6.2833]


Mathematica graphics






share|improve this answer









$endgroup$












  • $begingroup$
    Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
    $endgroup$
    – York Tsang
    2 hours ago













6












6








6





$begingroup$

I get the same result in Mathematica, so it's not a Mathematica Online issue. I don't think it's even a Mathematica issue. It's due to two factors:




  1. w1 is not equal to w2, because N doesn't actually truncate 2 Pi to five digits


  2. Det[k-w^2*m] changes quickly, so any little inaccuracy in w becomes a big discrepancy in Det[k-w^2*m]

To see #1:



w1 == 2 [Pi]
(* True *)
w1 - w2
(* -0.0000146928 *)


To see #2:



Plot[Det[k - w^2*m], w, 6.2831, 6.2833]


Mathematica graphics






share|improve this answer









$endgroup$



I get the same result in Mathematica, so it's not a Mathematica Online issue. I don't think it's even a Mathematica issue. It's due to two factors:




  1. w1 is not equal to w2, because N doesn't actually truncate 2 Pi to five digits


  2. Det[k-w^2*m] changes quickly, so any little inaccuracy in w becomes a big discrepancy in Det[k-w^2*m]

To see #1:



w1 == 2 [Pi]
(* True *)
w1 - w2
(* -0.0000146928 *)


To see #2:



Plot[Det[k - w^2*m], w, 6.2831, 6.2833]


Mathematica graphics







share|improve this answer












share|improve this answer



share|improve this answer










answered 8 hours ago









Chris KChris K

8,08722246




8,08722246











  • $begingroup$
    Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
    $endgroup$
    – York Tsang
    2 hours ago
















  • $begingroup$
    Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
    $endgroup$
    – York Tsang
    2 hours ago















$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
$endgroup$
– York Tsang
2 hours ago




$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ and N[2Pi,5] are not exactly the same. I have edited the question.
$endgroup$
– York Tsang
2 hours ago










York Tsang is a new contributor. Be nice, and check out our Code of Conduct.









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York Tsang is a new contributor. Be nice, and check out our Code of Conduct.












York Tsang is a new contributor. Be nice, and check out our Code of Conduct.











York Tsang is a new contributor. Be nice, and check out our Code of Conduct.














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