Plotting a Maclaurin series Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern) Announcing the arrival of Valued Associate #679: Cesar Manara Unicorn Meta Zoo #1: Why another podcast?Normal[Series[ ]] does not give a normal expressionPlot Series of Exp(x) about x=0 to order 12Plotting a few Maclaurin Series in MathematicaPlotting an infinite seriesPolynomial with alternating sign coefficients from the odd degree terms of a Truncated Power SeriesPuiseux series for algebraic curvesPlotting a Taylor series of Partial sumMaclaurin Series - TablePlotting a Taylor Series of two-variable trigonometric functionDefining Valuation over Puiseux SeriesPlotting an infinite seriesPlotting a few Maclaurin Series in MathematicaProblem plotting partial sum of a Fourier seriesPlotting an association key corresponding to the maximum value
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Plotting a Maclaurin series
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)
Announcing the arrival of Valued Associate #679: Cesar Manara
Unicorn Meta Zoo #1: Why another podcast?Normal[Series[ ]] does not give a normal expressionPlot Series of Exp(x) about x=0 to order 12Plotting a few Maclaurin Series in MathematicaPlotting an infinite seriesPolynomial with alternating sign coefficients from the odd degree terms of a Truncated Power SeriesPuiseux series for algebraic curvesPlotting a Taylor series of Partial sumMaclaurin Series - TablePlotting a Taylor Series of two-variable trigonometric functionDefining Valuation over Puiseux SeriesPlotting an infinite seriesPlotting a few Maclaurin Series in MathematicaProblem plotting partial sum of a Fourier seriesPlotting an association key corresponding to the maximum value
$begingroup$
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
New contributor
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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
New contributor
$endgroup$
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
1 hour 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
New contributor
$endgroup$
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
New contributor
New contributor
edited 3 hours ago
m_goldberg
89k873200
89k873200
New contributor
asked 3 hours ago
niknik
61
61
New contributor
New contributor
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
1 hour ago
add a comment |
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
1 hour ago
1
1
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
1 hour ago
$begingroup$
Possible duplicates: (7559), (130464), (158080), (160684)
$endgroup$
– Michael E2
1 hour ago
add a comment |
1 Answer
1
active
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]
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
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]
$endgroup$
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]
$endgroup$
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]
$endgroup$
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 3 hours 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
1 hour ago