f(x) having fixed significant digits in Table

Clash Royale CLAN TAG#URR8PPP

up vote
4
down vote

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Define function f

Clear[f, x];
f[x_] := 2^x;

N[ f[arg], #sig. digits]

data = Table[
 N[Pi, n],
 N[ f[N[Pi, n]], 10 ]
 , n, 1, 8
 ];

Text gridding...

Text@Grid[Prepend[data, "x", "f(x)"],
 Alignment -> Left,
 Dividers -> Center, 2 -> True
 ]

beginarrayl
textx & textf(x) \
hline
3. & 0. \
3.1 & 9. \
3.14 & 8.8 \
3.142 & 8.82 \
3.1416 & 8.825 \
3.14159 & 8.8250 \
3.141593 & 8.82498 \
3.1415927 & 8.824978 \
endarray

How do I show the following instead?

beginarrayl
textx & textf(x) \
hline
3. & 8.000000 \
3.1 & 8.574188 \
3.14 & 8.815241 \
3.142 & 8.821353 \
3.1416 & 8.824411 \
3.14159 & 8.824962 \
3.141593 & 8.824974 \
3.1415927 & 8.824978 \
endarray

table

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asked Sep 9 at 2:39

R5-D4

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up vote
4
down vote

favorite

Define function f

Clear[f, x];
f[x_] := 2^x;

N[ f[arg], #sig. digits]

data = Table[
 N[Pi, n],
 N[ f[N[Pi, n]], 10 ]
 , n, 1, 8
 ];

Text gridding...

Text@Grid[Prepend[data, "x", "f(x)"],
 Alignment -> Left,
 Dividers -> Center, 2 -> True
 ]

beginarrayl
textx & textf(x) \
hline
3. & 0. \
3.1 & 9. \
3.14 & 8.8 \
3.142 & 8.82 \
3.1416 & 8.825 \
3.14159 & 8.8250 \
3.141593 & 8.82498 \
3.1415927 & 8.824978 \
endarray

How do I show the following instead?

beginarrayl
textx & textf(x) \
hline
3. & 8.000000 \
3.1 & 8.574188 \
3.14 & 8.815241 \
3.142 & 8.821353 \
3.1416 & 8.824411 \
3.14159 & 8.824962 \
3.141593 & 8.824974 \
3.1415927 & 8.824978 \
endarray

table

share|improve this question

asked Sep 9 at 2:39

R5-D4

211

New contributor

R5-D4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

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up vote
4
down vote

favorite

up vote
4
down vote

favorite

Define function f

Clear[f, x];
f[x_] := 2^x;

N[ f[arg], #sig. digits]

data = Table[
 N[Pi, n],
 N[ f[N[Pi, n]], 10 ]
 , n, 1, 8
 ];

Text gridding...

Text@Grid[Prepend[data, "x", "f(x)"],
 Alignment -> Left,
 Dividers -> Center, 2 -> True
 ]

beginarrayl
textx & textf(x) \
hline
3. & 0. \
3.1 & 9. \
3.14 & 8.8 \
3.142 & 8.82 \
3.1416 & 8.825 \
3.14159 & 8.8250 \
3.141593 & 8.82498 \
3.1415927 & 8.824978 \
endarray

How do I show the following instead?

beginarrayl
textx & textf(x) \
hline
3. & 8.000000 \
3.1 & 8.574188 \
3.14 & 8.815241 \
3.142 & 8.821353 \
3.1416 & 8.824411 \
3.14159 & 8.824962 \
3.141593 & 8.824974 \
3.1415927 & 8.824978 \
endarray

table

share|improve this question

asked Sep 9 at 2:39

R5-D4

211

New contributor

R5-D4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

Define function f

Clear[f, x];
f[x_] := 2^x;

N[ f[arg], #sig. digits]

data = Table[
 N[Pi, n],
 N[ f[N[Pi, n]], 10 ]
 , n, 1, 8
 ];

Text gridding...

Text@Grid[Prepend[data, "x", "f(x)"],
 Alignment -> Left,
 Dividers -> Center, 2 -> True
 ]

beginarrayl
textx & textf(x) \
hline
3. & 0. \
3.1 & 9. \
3.14 & 8.8 \
3.142 & 8.82 \
3.1416 & 8.825 \
3.14159 & 8.8250 \
3.141593 & 8.82498 \
3.1415927 & 8.824978 \
endarray

How do I show the following instead?

beginarrayl
textx & textf(x) \
hline
3. & 8.000000 \
3.1 & 8.574188 \
3.14 & 8.815241 \
3.142 & 8.821353 \
3.1416 & 8.824411 \
3.14159 & 8.824962 \
3.141593 & 8.824974 \
3.1415927 & 8.824978 \
endarray

table

table

share|improve this question

asked Sep 9 at 2:39

R5-D4

211

New contributor

R5-D4 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

asked Sep 9 at 2:39

R5-D4

211

New contributor

R5-D4 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

asked Sep 9 at 2:39

R5-D4

211

New contributor

R5-D4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

asked Sep 9 at 2:39

R5-D4

211

asked Sep 9 at 2:39

R5-D4

211

211

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R5-D4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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New contributor

R5-D4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

R5-D4 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.

add a comment | 

add a comment | 

2 Answers
2

active

oldest

votes

up vote
6
down vote

Using arbitrary precision arithmetic produces the 1st result you show because Mathematica normally does not show digits with no precision since they are just noise. However, if you are trying to show how having better and better rational approximations to π improves the approximation of $2^pi$, I suggest the following approach.

data =
 With[x = π, n = 8,
 Table[
 With[u = Round[x, 10^-i], u, Round[f[u], 10^-i]] // N,
 i, 0, n]];
TableForm[Map[NumberForm[#, 10, 8] &, data, 2],
 TableHeadings -> None, x, f[x]]

share|improve this answer

answered Sep 9 at 4:13

m_goldberg

81.8k869189

add a comment | 

up vote
2
down vote

It might take some additional formatting to get exactly what you're after, but this is a start. I acknowledge some things may be able to be made simpler.

Clear[f, x];
f[x_] := 2^x;
data = N@Table[
 Table[10^(-j + 1), j, 1, i].RealDigits[N@Pi, 10, i][[1]], i, 
 8] /. x_?NumericQ :> NumberForm[x, 8], f[x];

Text@Grid[Prepend[data, "x", "f(x)"], Alignment -> Left, 
 Dividers -> Center, 2 -> True]

share|improve this answer

answered Sep 9 at 2:56

user6014

2,5721021

add a comment | 

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
6
down vote

Using arbitrary precision arithmetic produces the 1st result you show because Mathematica normally does not show digits with no precision since they are just noise. However, if you are trying to show how having better and better rational approximations to π improves the approximation of $2^pi$, I suggest the following approach.

data =
 With[x = π, n = 8,
 Table[
 With[u = Round[x, 10^-i], u, Round[f[u], 10^-i]] // N,
 i, 0, n]];
TableForm[Map[NumberForm[#, 10, 8] &, data, 2],
 TableHeadings -> None, x, f[x]]

share|improve this answer

answered Sep 9 at 4:13

m_goldberg

81.8k869189

add a comment | 

up vote
6
down vote

Using arbitrary precision arithmetic produces the 1st result you show because Mathematica normally does not show digits with no precision since they are just noise. However, if you are trying to show how having better and better rational approximations to π improves the approximation of $2^pi$, I suggest the following approach.

data =
 With[x = π, n = 8,
 Table[
 With[u = Round[x, 10^-i], u, Round[f[u], 10^-i]] // N,
 i, 0, n]];
TableForm[Map[NumberForm[#, 10, 8] &, data, 2],
 TableHeadings -> None, x, f[x]]

share|improve this answer

answered Sep 9 at 4:13

m_goldberg

81.8k869189

add a comment | 

up vote
6
down vote

up vote
6
down vote

Using arbitrary precision arithmetic produces the 1st result you show because Mathematica normally does not show digits with no precision since they are just noise. However, if you are trying to show how having better and better rational approximations to π improves the approximation of $2^pi$, I suggest the following approach.

data =
 With[x = π, n = 8,
 Table[
 With[u = Round[x, 10^-i], u, Round[f[u], 10^-i]] // N,
 i, 0, n]];
TableForm[Map[NumberForm[#, 10, 8] &, data, 2],
 TableHeadings -> None, x, f[x]]

share|improve this answer

answered Sep 9 at 4:13

m_goldberg

81.8k869189

Using arbitrary precision arithmetic produces the 1st result you show because Mathematica normally does not show digits with no precision since they are just noise. However, if you are trying to show how having better and better rational approximations to π improves the approximation of $2^pi$, I suggest the following approach.

data =
 With[x = π, n = 8,
 Table[
 With[u = Round[x, 10^-i], u, Round[f[u], 10^-i]] // N,
 i, 0, n]];
TableForm[Map[NumberForm[#, 10, 8] &, data, 2],
 TableHeadings -> None, x, f[x]]

share|improve this answer

answered Sep 9 at 4:13

m_goldberg

81.8k869189

share|improve this answer

share|improve this answer

answered Sep 9 at 4:13

m_goldberg

81.8k869189

answered Sep 9 at 4:13

m_goldberg

81.8k869189

answered Sep 9 at 4:13

m_goldberg

81.8k869189

81.8k869189

add a comment | 

add a comment | 

up vote
2
down vote

It might take some additional formatting to get exactly what you're after, but this is a start. I acknowledge some things may be able to be made simpler.

Clear[f, x];
f[x_] := 2^x;
data = N@Table[
 Table[10^(-j + 1), j, 1, i].RealDigits[N@Pi, 10, i][[1]], i, 
 8] /. x_?NumericQ :> NumberForm[x, 8], f[x];

Text@Grid[Prepend[data, "x", "f(x)"], Alignment -> Left, 
 Dividers -> Center, 2 -> True]

share|improve this answer

answered Sep 9 at 2:56

user6014

2,5721021

add a comment | 

up vote
2
down vote

It might take some additional formatting to get exactly what you're after, but this is a start. I acknowledge some things may be able to be made simpler.

Clear[f, x];
f[x_] := 2^x;
data = N@Table[
 Table[10^(-j + 1), j, 1, i].RealDigits[N@Pi, 10, i][[1]], i, 
 8] /. x_?NumericQ :> NumberForm[x, 8], f[x];

Text@Grid[Prepend[data, "x", "f(x)"], Alignment -> Left, 
 Dividers -> Center, 2 -> True]

share|improve this answer

answered Sep 9 at 2:56

user6014

2,5721021

add a comment | 

up vote
2
down vote

up vote
2
down vote

It might take some additional formatting to get exactly what you're after, but this is a start. I acknowledge some things may be able to be made simpler.

Clear[f, x];
f[x_] := 2^x;
data = N@Table[
 Table[10^(-j + 1), j, 1, i].RealDigits[N@Pi, 10, i][[1]], i, 
 8] /. x_?NumericQ :> NumberForm[x, 8], f[x];

Text@Grid[Prepend[data, "x", "f(x)"], Alignment -> Left, 
 Dividers -> Center, 2 -> True]

share|improve this answer

answered Sep 9 at 2:56

user6014

2,5721021

It might take some additional formatting to get exactly what you're after, but this is a start. I acknowledge some things may be able to be made simpler.

Clear[f, x];
f[x_] := 2^x;
data = N@Table[
 Table[10^(-j + 1), j, 1, i].RealDigits[N@Pi, 10, i][[1]], i, 
 8] /. x_?NumericQ :> NumberForm[x, 8], f[x];

Text@Grid[Prepend[data, "x", "f(x)"], Alignment -> Left, 
 Dividers -> Center, 2 -> True]

share|improve this answer

answered Sep 9 at 2:56

user6014

2,5721021

share|improve this answer

share|improve this answer

answered Sep 9 at 2:56

user6014

2,5721021

answered Sep 9 at 2:56

user6014

2,5721021

answered Sep 9 at 2:56

user6014

2,5721021

2,5721021

add a comment | 

add a comment | 

R5-D4 is a new contributor. Be nice, and check out our Code of Conduct.

 

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