Inconsistent results from Wolfram CouldFind 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
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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
$begingroup$
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:
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:
So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?
linear-algebra
asked 8 hours ago
York TsangYork Tsang
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.
$endgroup$
add a comment |
$begingroup$
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:
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:
So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?
linear-algebra
asked 8 hours ago
York TsangYork Tsang
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.
$endgroup$
3
$begingroup$
Suppose small epsilon thenClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]]returns12500000*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
7 hours ago
2
$begingroup$
No,N[x, p], represents, if possible, the value ofxapproximated to a precision ofpdigits. Read the documentation onN.
$endgroup$
– 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
add a comment |
$begingroup$
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:
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:
So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?
linear-algebra
asked 8 hours ago
York TsangYork Tsang
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.
$endgroup$
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:
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:
So, is N[2Pi,5] exactly equal to $2pi$ or not? What does the function N actually do?
linear-algebra
linear-algebra
asked 8 hours ago
York TsangYork Tsang
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.
asked 8 hours ago
York TsangYork Tsang
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.
edited 2 hours ago
York Tsang
asked 8 hours ago
York TsangYork Tsang
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.
asked 8 hours ago
York TsangYork Tsang
1063
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 thenClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]]returns12500000*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
7 hours ago
2
$begingroup$
No,N[x, p], represents, if possible, the value ofxapproximated to a precision ofpdigits. Read the documentation onN.
$endgroup$
– 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
add a comment |
3
$begingroup$
Suppose small epsilon thenClearAll["Global`*"]; m=2,0,0,1*2500; k=3,-1,-1,1*20000 Pi^2; w1=2Pi+epsilon; FullSimplify[Det[k-w1^2*m]]returns12500000*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
7 hours ago
2
$begingroup$
No,N[x, p], represents, if possible, the value ofxapproximated to a precision ofpdigits. Read the documentation onN.
$endgroup$
– 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
7 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
7 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
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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:
w1is not equal tow2, becauseNdoesn't actually truncate2 Pito five digitsDet[k-w^2*m]changes quickly, so any little inaccuracy inwbecomes a big discrepancy inDet[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]
answered 7 hours ago
Chris KChris K
8,08722246
$endgroup$
$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ andN[2Pi,5]are not exactly the same. I have edited the question.
$endgroup$
– York Tsang
2 hours ago
add a comment |
Your Answer
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1 Answer
1
active
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1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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:
w1is not equal tow2, becauseNdoesn't actually truncate2 Pito five digitsDet[k-w^2*m]changes quickly, so any little inaccuracy inwbecomes a big discrepancy inDet[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]
answered 7 hours ago
Chris KChris K
8,08722246
$endgroup$
$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ andN[2Pi,5]are not exactly the same. I have edited the question.
$endgroup$
– York Tsang
2 hours ago
add a comment |
$begingroup$
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:
w1is not equal tow2, becauseNdoesn't actually truncate2 Pito five digitsDet[k-w^2*m]changes quickly, so any little inaccuracy inwbecomes a big discrepancy inDet[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]
answered 7 hours ago
Chris KChris K
8,08722246
$endgroup$
$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ andN[2Pi,5]are not exactly the same. I have edited the question.
$endgroup$
– York Tsang
2 hours ago
add a comment |
$begingroup$
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:
w1is not equal tow2, becauseNdoesn't actually truncate2 Pito five digitsDet[k-w^2*m]changes quickly, so any little inaccuracy inwbecomes a big discrepancy inDet[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]
answered 7 hours ago
Chris KChris K
8,08722246
$endgroup$
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:
w1is not equal tow2, becauseNdoesn't actually truncate2 Pito five digitsDet[k-w^2*m]changes quickly, so any little inaccuracy inwbecomes a big discrepancy inDet[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]
answered 7 hours ago
Chris KChris K
8,08722246
answered 7 hours ago
Chris KChris K
8,08722246
answered 7 hours ago
Chris KChris K
8,08722246
answered 7 hours ago
Chris KChris K
8,08722246
8,08722246
$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ andN[2Pi,5]are not exactly the same. I have edited the question.
$endgroup$
– York Tsang
2 hours ago
add a comment |
$begingroup$
Regarding #1, it appears that the determinants calculated using $2pi$ andN[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
add a comment |
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.
York Tsang is a new contributor. Be nice, and check out our Code of Conduct.
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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]]returns12500000*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
7 hours ago
2
$begingroup$
No,
N[x, p], represents, if possible, the value ofxapproximated to a precision ofpdigits. Read the documentation onN.$endgroup$
– 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