How to display very small numbers in Mathematica?
$begingroup$
I am trying to evaluate the function:
$$f(x) = cos(x) - mathrm{e}^{-2.7 x}$$
at $x = 1.7 times 10^{-25}$
and Mathematica keeps returning '0.'
How do I evaluate the expression in a better way?
precision-and-accuracy
edited 18 mins ago
Henrik Schumacher
50.8k469145
asked 3 hours ago
Ray_56Ray_56
111
New contributor
Ray_56 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 |
$begingroup$
I am trying to evaluate the function:
$$f(x) = cos(x) - mathrm{e}^{-2.7 x}$$
at $x = 1.7 times 10^{-25}$
and Mathematica keeps returning '0.'
How do I evaluate the expression in a better way?
precision-and-accuracy
edited 18 mins ago
Henrik Schumacher
50.8k469145
asked 3 hours ago
Ray_56Ray_56
111
New contributor
Ray_56 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 |
$begingroup$
I am trying to evaluate the function:
$$f(x) = cos(x) - mathrm{e}^{-2.7 x}$$
at $x = 1.7 times 10^{-25}$
and Mathematica keeps returning '0.'
How do I evaluate the expression in a better way?
precision-and-accuracy
edited 18 mins ago
Henrik Schumacher
50.8k469145
asked 3 hours ago
Ray_56Ray_56
111
New contributor
Ray_56 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
I am trying to evaluate the function:
$$f(x) = cos(x) - mathrm{e}^{-2.7 x}$$
at $x = 1.7 times 10^{-25}$
and Mathematica keeps returning '0.'
How do I evaluate the expression in a better way?
precision-and-accuracy
precision-and-accuracy
edited 18 mins ago
Henrik Schumacher
50.8k469145
asked 3 hours ago
Ray_56Ray_56
111
New contributor
Ray_56 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 mins ago
Henrik Schumacher
50.8k469145
asked 3 hours ago
Ray_56Ray_56
111
New contributor
Ray_56 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 18 mins ago
Henrik Schumacher
50.8k469145
edited 18 mins ago
Henrik Schumacher
50.8k469145
edited 18 mins ago
Henrik Schumacher
50.8k469145
50.8k469145
asked 3 hours ago
Ray_56Ray_56
111
New contributor
Ray_56 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 3 hours ago
Ray_56Ray_56
111
asked 3 hours ago
Ray_56Ray_56
111
111
New contributor
Ray_56 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Ray_56 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Ray_56 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 |
4 Answers
4
active
oldest
votes
$begingroup$
Can also use exact numbers.
f[x_] = Cos[x] - E^(-27 x/10);
f[17 10^-26]//N[#,50]&
(*4.5899999999999999999999988020950000000000000000002*10^-25*)
answered 1 hour ago
Bill WattsBill Watts
3,0811518
$endgroup$
add a comment |
$begingroup$
The simplest methods are usually the best. Try this code
N[Cos[x] - Exp[-x 27/10] /. x -> 17*^-26, 15] // InputForm
or a minor variation
With[{x = 17*^-26}, N[Cos[x] - Exp[-x 27/10], 15]] // InputForm
or define a function first
f = Cos[#] - Exp[-# 27/10] &; N[f[17*^-26], 15] // InputForm
All of these return the result
4.589999999999999999999998802094999`15.*^-25
You can get more digits by increasing the 15 digits precision.
For example, with 34 digits precision the result returned is
4.5899999999999999999999988020950000000000000000001613`34.*^-25
answered 50 mins ago
SomosSomos
4778
$endgroup$
add a comment |
$begingroup$
First convert the expression to trigonometric form:
y = Cos[x] - Exp[-2.7*x] // ExpToTrig
Cos[x] - Cosh[2.7 x] + Sinh[2.7 x]
y /. x -> 1.7*10^-25
4.59*10^-25
This is the exact solution.
edited 16 mins ago
Henrik Schumacher
50.8k469145
answered 1 hour ago
VixillatorVixillator
4497
$endgroup$
add a comment |
$begingroup$
Do a series expansion:
Series[Cos[x] - Exp[-2.7 x], {x, 0, 1}]
(*
SeriesData[x, 0, {2.7}, 1, 2, 1]
*)
Then plug in $x = 1.7 times 10^{-25}$ to get:
$$4.59 times 10^{-25}$$
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Can also use exact numbers.
f[x_] = Cos[x] - E^(-27 x/10);
f[17 10^-26]//N[#,50]&
(*4.5899999999999999999999988020950000000000000000002*10^-25*)
answered 1 hour ago
Bill WattsBill Watts
3,0811518
$endgroup$
add a comment |
$begingroup$
Can also use exact numbers.
f[x_] = Cos[x] - E^(-27 x/10);
f[17 10^-26]//N[#,50]&
(*4.5899999999999999999999988020950000000000000000002*10^-25*)
answered 1 hour ago
Bill WattsBill Watts
3,0811518
$endgroup$
add a comment |
$begingroup$
Can also use exact numbers.
f[x_] = Cos[x] - E^(-27 x/10);
f[17 10^-26]//N[#,50]&
(*4.5899999999999999999999988020950000000000000000002*10^-25*)
answered 1 hour ago
Bill WattsBill Watts
3,0811518
$endgroup$
Can also use exact numbers.
f[x_] = Cos[x] - E^(-27 x/10);
f[17 10^-26]//N[#,50]&
(*4.5899999999999999999999988020950000000000000000002*10^-25*)
answered 1 hour ago
Bill WattsBill Watts
3,0811518
answered 1 hour ago
Bill WattsBill Watts
3,0811518
answered 1 hour ago
Bill WattsBill Watts
3,0811518
answered 1 hour ago
Bill WattsBill Watts
3,0811518
3,0811518
add a comment |
add a comment |
$begingroup$
The simplest methods are usually the best. Try this code
N[Cos[x] - Exp[-x 27/10] /. x -> 17*^-26, 15] // InputForm
or a minor variation
With[{x = 17*^-26}, N[Cos[x] - Exp[-x 27/10], 15]] // InputForm
or define a function first
f = Cos[#] - Exp[-# 27/10] &; N[f[17*^-26], 15] // InputForm
All of these return the result
4.589999999999999999999998802094999`15.*^-25
You can get more digits by increasing the 15 digits precision.
For example, with 34 digits precision the result returned is
4.5899999999999999999999988020950000000000000000001613`34.*^-25
answered 50 mins ago
SomosSomos
4778
$endgroup$
add a comment |
$begingroup$
The simplest methods are usually the best. Try this code
N[Cos[x] - Exp[-x 27/10] /. x -> 17*^-26, 15] // InputForm
or a minor variation
With[{x = 17*^-26}, N[Cos[x] - Exp[-x 27/10], 15]] // InputForm
or define a function first
f = Cos[#] - Exp[-# 27/10] &; N[f[17*^-26], 15] // InputForm
All of these return the result
4.589999999999999999999998802094999`15.*^-25
You can get more digits by increasing the 15 digits precision.
For example, with 34 digits precision the result returned is
4.5899999999999999999999988020950000000000000000001613`34.*^-25
answered 50 mins ago
SomosSomos
4778
$endgroup$
add a comment |
$begingroup$
The simplest methods are usually the best. Try this code
N[Cos[x] - Exp[-x 27/10] /. x -> 17*^-26, 15] // InputForm
or a minor variation
With[{x = 17*^-26}, N[Cos[x] - Exp[-x 27/10], 15]] // InputForm
or define a function first
f = Cos[#] - Exp[-# 27/10] &; N[f[17*^-26], 15] // InputForm
All of these return the result
4.589999999999999999999998802094999`15.*^-25
You can get more digits by increasing the 15 digits precision.
For example, with 34 digits precision the result returned is
4.5899999999999999999999988020950000000000000000001613`34.*^-25
answered 50 mins ago
SomosSomos
4778
$endgroup$
The simplest methods are usually the best. Try this code
N[Cos[x] - Exp[-x 27/10] /. x -> 17*^-26, 15] // InputForm
or a minor variation
With[{x = 17*^-26}, N[Cos[x] - Exp[-x 27/10], 15]] // InputForm
or define a function first
f = Cos[#] - Exp[-# 27/10] &; N[f[17*^-26], 15] // InputForm
All of these return the result
4.589999999999999999999998802094999`15.*^-25
You can get more digits by increasing the 15 digits precision.
For example, with 34 digits precision the result returned is
4.5899999999999999999999988020950000000000000000001613`34.*^-25
answered 50 mins ago
SomosSomos
4778
edited 34 mins ago
answered 50 mins ago
SomosSomos
4778
answered 50 mins ago
SomosSomos
4778
answered 50 mins ago
SomosSomos
4778
4778
add a comment |
add a comment |
$begingroup$
First convert the expression to trigonometric form:
y = Cos[x] - Exp[-2.7*x] // ExpToTrig
Cos[x] - Cosh[2.7 x] + Sinh[2.7 x]
y /. x -> 1.7*10^-25
4.59*10^-25
This is the exact solution.
edited 16 mins ago
Henrik Schumacher
50.8k469145
answered 1 hour ago
VixillatorVixillator
4497
$endgroup$
add a comment |
$begingroup$
First convert the expression to trigonometric form:
y = Cos[x] - Exp[-2.7*x] // ExpToTrig
Cos[x] - Cosh[2.7 x] + Sinh[2.7 x]
y /. x -> 1.7*10^-25
4.59*10^-25
This is the exact solution.
edited 16 mins ago
Henrik Schumacher
50.8k469145
answered 1 hour ago
VixillatorVixillator
4497
$endgroup$
add a comment |
$begingroup$
First convert the expression to trigonometric form:
y = Cos[x] - Exp[-2.7*x] // ExpToTrig
Cos[x] - Cosh[2.7 x] + Sinh[2.7 x]
y /. x -> 1.7*10^-25
4.59*10^-25
This is the exact solution.
edited 16 mins ago
Henrik Schumacher
50.8k469145
answered 1 hour ago
VixillatorVixillator
4497
$endgroup$
First convert the expression to trigonometric form:
y = Cos[x] - Exp[-2.7*x] // ExpToTrig
Cos[x] - Cosh[2.7 x] + Sinh[2.7 x]
y /. x -> 1.7*10^-25
4.59*10^-25
This is the exact solution.
edited 16 mins ago
Henrik Schumacher
50.8k469145
answered 1 hour ago
VixillatorVixillator
4497
edited 16 mins ago
Henrik Schumacher
50.8k469145
edited 16 mins ago
Henrik Schumacher
50.8k469145
edited 16 mins ago
Henrik Schumacher
50.8k469145
50.8k469145
answered 1 hour ago
VixillatorVixillator
4497
answered 1 hour ago
VixillatorVixillator
4497
answered 1 hour ago
VixillatorVixillator
4497
4497
add a comment |
add a comment |
$begingroup$
Do a series expansion:
Series[Cos[x] - Exp[-2.7 x], {x, 0, 1}]
(*
SeriesData[x, 0, {2.7}, 1, 2, 1]
*)
Then plug in $x = 1.7 times 10^{-25}$ to get:
$$4.59 times 10^{-25}$$
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
add a comment |
$begingroup$
Do a series expansion:
Series[Cos[x] - Exp[-2.7 x], {x, 0, 1}]
(*
SeriesData[x, 0, {2.7}, 1, 2, 1]
*)
Then plug in $x = 1.7 times 10^{-25}$ to get:
$$4.59 times 10^{-25}$$
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
add a comment |
$begingroup$
Do a series expansion:
Series[Cos[x] - Exp[-2.7 x], {x, 0, 1}]
(*
SeriesData[x, 0, {2.7}, 1, 2, 1]
*)
Then plug in $x = 1.7 times 10^{-25}$ to get:
$$4.59 times 10^{-25}$$
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
$endgroup$
Do a series expansion:
Series[Cos[x] - Exp[-2.7 x], {x, 0, 1}]
(*
SeriesData[x, 0, {2.7}, 1, 2, 1]
*)
Then plug in $x = 1.7 times 10^{-25}$ to get:
$$4.59 times 10^{-25}$$
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
answered 2 hours ago
David G. StorkDavid G. Stork
23.9k22153
23.9k22153
add a comment |
add a comment |
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