Using Maple 2018 to find coefficients in a power series
I'm doing a practical experiement and have measured the voltage over a capacitor on an oscilloscope as the capacitor decays.
I have a table of results showing the voltage and time, and created an exponential fit and a plot, using
withStatistics
entering the x,y co ordinates
ExponentialFit(X, Y, v)
This calculates $Ae^x$ but I would like to get the first four coefficients of the power series for my measured value of $e^x$.
Any help on how to do this would be much appreciated!
power-series exponential-function maple
asked Nov 29 '18 at 20:17
WilliamWilliam
52
add a comment |
I'm doing a practical experiement and have measured the voltage over a capacitor on an oscilloscope as the capacitor decays.
I have a table of results showing the voltage and time, and created an exponential fit and a plot, using
withStatistics
entering the x,y co ordinates
ExponentialFit(X, Y, v)
This calculates $Ae^x$ but I would like to get the first four coefficients of the power series for my measured value of $e^x$.
Any help on how to do this would be much appreciated!
power-series exponential-function maple
asked Nov 29 '18 at 20:17
WilliamWilliam
52
add a comment |
I'm doing a practical experiement and have measured the voltage over a capacitor on an oscilloscope as the capacitor decays.
I have a table of results showing the voltage and time, and created an exponential fit and a plot, using
withStatistics
entering the x,y co ordinates
ExponentialFit(X, Y, v)
This calculates $Ae^x$ but I would like to get the first four coefficients of the power series for my measured value of $e^x$.
Any help on how to do this would be much appreciated!
power-series exponential-function maple
asked Nov 29 '18 at 20:17
WilliamWilliam
52
I'm doing a practical experiement and have measured the voltage over a capacitor on an oscilloscope as the capacitor decays.
I have a table of results showing the voltage and time, and created an exponential fit and a plot, using
withStatistics
entering the x,y co ordinates
ExponentialFit(X, Y, v)
This calculates $Ae^x$ but I would like to get the first four coefficients of the power series for my measured value of $e^x$.
Any help on how to do this would be much appreciated!
power-series exponential-function maple
power-series exponential-function maple
asked Nov 29 '18 at 20:17
WilliamWilliam
52
asked Nov 29 '18 at 20:17
WilliamWilliam
52
edited Nov 29 '18 at 20:31
William
asked Nov 29 '18 at 20:17
WilliamWilliam
52
asked Nov 29 '18 at 20:17
WilliamWilliam
52
asked Nov 29 '18 at 20:17
WilliamWilliam
52
52
add a comment |
add a comment |
1 Answer
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So, if I understand, you've found a numeric value for A that makes A*exp(x) fit the data.
If so, then is there a problem with substituting the parameter value into the power series for A*exp(x)?
raw := convert(series(A*exp(x), x, 4), polynom):
lprint(raw);
A+A*x+(1/2)*A*x^2+(1/6)*A*x^3
eval(raw, A=2.134); # or what have you
2 3
2.134 + 2.134 x + 1.067000000 x + 0.3556666667 x
Or, if you already have an expression with the numeric value,
expr := 2.134*exp(x):
raw := convert(series(expr, x, 4), polynom):
lprint(raw);
2.134+2.134*x+1.067000000*x^2+.3556666667*x^3
Of course, for this example that is just A multiplied by the series for exp(x).
answered Nov 29 '18 at 22:19
aceracer
3,640199
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
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So, if I understand, you've found a numeric value for A that makes A*exp(x) fit the data.
If so, then is there a problem with substituting the parameter value into the power series for A*exp(x)?
raw := convert(series(A*exp(x), x, 4), polynom):
lprint(raw);
A+A*x+(1/2)*A*x^2+(1/6)*A*x^3
eval(raw, A=2.134); # or what have you
2 3
2.134 + 2.134 x + 1.067000000 x + 0.3556666667 x
Or, if you already have an expression with the numeric value,
expr := 2.134*exp(x):
raw := convert(series(expr, x, 4), polynom):
lprint(raw);
2.134+2.134*x+1.067000000*x^2+.3556666667*x^3
Of course, for this example that is just A multiplied by the series for exp(x).
answered Nov 29 '18 at 22:19
aceracer
3,640199
add a comment |
So, if I understand, you've found a numeric value for A that makes A*exp(x) fit the data.
If so, then is there a problem with substituting the parameter value into the power series for A*exp(x)?
raw := convert(series(A*exp(x), x, 4), polynom):
lprint(raw);
A+A*x+(1/2)*A*x^2+(1/6)*A*x^3
eval(raw, A=2.134); # or what have you
2 3
2.134 + 2.134 x + 1.067000000 x + 0.3556666667 x
Or, if you already have an expression with the numeric value,
expr := 2.134*exp(x):
raw := convert(series(expr, x, 4), polynom):
lprint(raw);
2.134+2.134*x+1.067000000*x^2+.3556666667*x^3
Of course, for this example that is just A multiplied by the series for exp(x).
answered Nov 29 '18 at 22:19
aceracer
3,640199
add a comment |
So, if I understand, you've found a numeric value for A that makes A*exp(x) fit the data.
If so, then is there a problem with substituting the parameter value into the power series for A*exp(x)?
raw := convert(series(A*exp(x), x, 4), polynom):
lprint(raw);
A+A*x+(1/2)*A*x^2+(1/6)*A*x^3
eval(raw, A=2.134); # or what have you
2 3
2.134 + 2.134 x + 1.067000000 x + 0.3556666667 x
Or, if you already have an expression with the numeric value,
expr := 2.134*exp(x):
raw := convert(series(expr, x, 4), polynom):
lprint(raw);
2.134+2.134*x+1.067000000*x^2+.3556666667*x^3
Of course, for this example that is just A multiplied by the series for exp(x).
answered Nov 29 '18 at 22:19
aceracer
3,640199
So, if I understand, you've found a numeric value for A that makes A*exp(x) fit the data.
If so, then is there a problem with substituting the parameter value into the power series for A*exp(x)?
raw := convert(series(A*exp(x), x, 4), polynom):
lprint(raw);
A+A*x+(1/2)*A*x^2+(1/6)*A*x^3
eval(raw, A=2.134); # or what have you
2 3
2.134 + 2.134 x + 1.067000000 x + 0.3556666667 x
Or, if you already have an expression with the numeric value,
expr := 2.134*exp(x):
raw := convert(series(expr, x, 4), polynom):
lprint(raw);
2.134+2.134*x+1.067000000*x^2+.3556666667*x^3
Of course, for this example that is just A multiplied by the series for exp(x).
answered Nov 29 '18 at 22:19
aceracer
3,640199
answered Nov 29 '18 at 22:19
aceracer
3,640199
answered Nov 29 '18 at 22:19
aceracer
3,640199
answered Nov 29 '18 at 22:19
aceracer
3,640199
3,640199
add a comment |
add a comment |
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