Plotting a Maclaurin series Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm...
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Plotting a Maclaurin series
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
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Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.
Please help, I've tried everything I could think of.
plotting series-expansion
edited 1 hour ago
m_goldberg
89k873200
asked 2 hours ago
niknik
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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$endgroup$
add a comment |
$begingroup$
Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.
Please help, I've tried everything I could think of.
plotting series-expansion
edited 1 hour ago
m_goldberg
89k873200
asked 2 hours ago
niknik
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
28 mins ago
add a comment |
$begingroup$
Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.
Please help, I've tried everything I could think of.
plotting series-expansion
edited 1 hour ago
m_goldberg
89k873200
asked 2 hours ago
niknik
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Can anyone plot the degree-8 Maclaurin polynomial for the function $(x+1)/((x^2)-9)$ and function $(x+1)/((x^2)-9)$ on the same plot.
Please help, I've tried everything I could think of.
plotting series-expansion
plotting series-expansion
edited 1 hour ago
m_goldberg
89k873200
asked 2 hours ago
niknik
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
m_goldberg
89k873200
asked 2 hours ago
niknik
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
m_goldberg
89k873200
edited 1 hour ago
m_goldberg
89k873200
edited 1 hour ago
m_goldberg
89k873200
89k873200
asked 2 hours ago
niknik
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
niknik
61
asked 2 hours ago
niknik
61
61
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
nik is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
28 mins ago
add a comment |
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
28 mins ago
1
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
28 mins ago
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
28 mins ago
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.
f = (x + 1)/((x^2) - 9);
p = Normal[Series[f, {x, 0, 8}]]
Plot[{f, p}, {x, -2, 2}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
add a comment |
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1 Answer
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active
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1 Answer
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active
oldest
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active
oldest
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votes
$begingroup$
The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.
f = (x + 1)/((x^2) - 9);
p = Normal[Series[f, {x, 0, 8}]]
Plot[{f, p}, {x, -2, 2}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
add a comment |
$begingroup$
The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.
f = (x + 1)/((x^2) - 9);
p = Normal[Series[f, {x, 0, 8}]]
Plot[{f, p}, {x, -2, 2}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
add a comment |
$begingroup$
The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.
f = (x + 1)/((x^2) - 9);
p = Normal[Series[f, {x, 0, 8}]]
Plot[{f, p}, {x, -2, 2}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
$endgroup$
The problem is probably that Series returns a SeriesData object. It must be converted to a polynomial with Normal before it can be plotted.
f = (x + 1)/((x^2) - 9);
p = Normal[Series[f, {x, 0, 8}]]
Plot[{f, p}, {x, -2, 2}]
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
answered 1 hour ago
Henrik SchumacherHenrik Schumacher
60.7k585171
60.7k585171
add a comment |
add a comment |
nik is a new contributor. Be nice, and check out our Code of Conduct.
nik is a new contributor. Be nice, and check out our Code of Conduct.
nik is a new contributor. Be nice, and check out our Code of Conduct.
nik is a new contributor. Be nice, and check out our Code of Conduct.
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1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
28 mins ago