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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:

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

share|improve this question

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.

$endgroup$

• 
  
  
  

  
  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
  7 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
  

  
  
  
  

add a comment | 

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:

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

share|improve this question

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.

$endgroup$

• 
  
  
  

  
  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
  7 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
  

  
  
  
  

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

share|improve this question

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.

$endgroup$

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]

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]

share|improve this question

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.

share|improve this question

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

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

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]

share|improve this answer

answered 7 hours ago

Chris KChris K

8,08722246

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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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]

share|improve this answer

answered 7 hours ago

Chris KChris K

8,08722246

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

add a comment | 

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]

share|improve this answer

answered 7 hours ago

Chris KChris K

8,08722246

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  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
  

  
  
  
  

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:

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]

share|improve this answer

answered 7 hours ago

Chris KChris K

8,08722246

$endgroup$

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

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

share|improve this answer

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