Mixing
Taking derivatives of a power series
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I've been working on understanding power series, and came across a problem asking for the derivative of a certain power series and for the derivative to be a summation with a lower limit equal to zero.
Here is the summation:
$$sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}$$
I tried to write out the sum as terms and then took the derivative to get:
$$1-frac12x^3+frac7{512}x^6-frac{10}{110592}x^9+...$$
I am not sure if rewriting this as a summation will give me the answer, but if it will, I do not know how to do this.
calculus derivatives summation power-series
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
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add a comment |
$begingroup$
I've been working on understanding power series, and came across a problem asking for the derivative of a certain power series and for the derivative to be a summation with a lower limit equal to zero.
Here is the summation:
$$sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}$$
I tried to write out the sum as terms and then took the derivative to get:
$$1-frac12x^3+frac7{512}x^6-frac{10}{110592}x^9+...$$
I am not sure if rewriting this as a summation will give me the answer, but if it will, I do not know how to do this.
calculus derivatives summation power-series
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
$endgroup$
$begingroup$
Assuming that $(n!)^3$ is a constant, the summation is $frac {x}{(n!)^3}sum_{k=0}^{infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $frac {x}{(n!)^3}frac {1}{1+x^3/2^3}$.
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– DanielWainfleet
Jan 10 at 1:03
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Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything.
$endgroup$
– Claude Leibovici
Jan 10 at 7:10
$begingroup$
Thanks for catching that, it was a typo, and it should be fixed now.
$endgroup$
– AspiringRoboteer
Jan 10 at 13:29
add a comment |
$begingroup$
I've been working on understanding power series, and came across a problem asking for the derivative of a certain power series and for the derivative to be a summation with a lower limit equal to zero.
Here is the summation:
$$sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}$$
I tried to write out the sum as terms and then took the derivative to get:
$$1-frac12x^3+frac7{512}x^6-frac{10}{110592}x^9+...$$
I am not sure if rewriting this as a summation will give me the answer, but if it will, I do not know how to do this.
calculus derivatives summation power-series
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
$endgroup$
I've been working on understanding power series, and came across a problem asking for the derivative of a certain power series and for the derivative to be a summation with a lower limit equal to zero.
Here is the summation:
$$sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}$$
I tried to write out the sum as terms and then took the derivative to get:
$$1-frac12x^3+frac7{512}x^6-frac{10}{110592}x^9+...$$
I am not sure if rewriting this as a summation will give me the answer, but if it will, I do not know how to do this.
calculus derivatives summation power-series
calculus derivatives summation power-series
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
edited Jan 10 at 13:29
AspiringRoboteer
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
asked Jan 9 at 21:09
AspiringRoboteerAspiringRoboteer
32
32
$begingroup$
Assuming that $(n!)^3$ is a constant, the summation is $frac {x}{(n!)^3}sum_{k=0}^{infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $frac {x}{(n!)^3}frac {1}{1+x^3/2^3}$.
$endgroup$
– DanielWainfleet
Jan 10 at 1:03
$begingroup$
Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything.
$endgroup$
– Claude Leibovici
Jan 10 at 7:10
$begingroup$
Thanks for catching that, it was a typo, and it should be fixed now.
$endgroup$
– AspiringRoboteer
Jan 10 at 13:29
add a comment |
$begingroup$
Assuming that $(n!)^3$ is a constant, the summation is $frac {x}{(n!)^3}sum_{k=0}^{infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $frac {x}{(n!)^3}frac {1}{1+x^3/2^3}$.
$endgroup$
– DanielWainfleet
Jan 10 at 1:03
$begingroup$
Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything.
$endgroup$
– Claude Leibovici
Jan 10 at 7:10
$begingroup$
Thanks for catching that, it was a typo, and it should be fixed now.
$endgroup$
– AspiringRoboteer
Jan 10 at 13:29
$begingroup$
Assuming that $(n!)^3$ is a constant, the summation is $frac {x}{(n!)^3}sum_{k=0}^{infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $frac {x}{(n!)^3}frac {1}{1+x^3/2^3}$.
$endgroup$
– DanielWainfleet
Jan 10 at 1:03
$begingroup$
Assuming that $(n!)^3$ is a constant, the summation is $frac {x}{(n!)^3}sum_{k=0}^{infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $frac {x}{(n!)^3}frac {1}{1+x^3/2^3}$.
$endgroup$
– DanielWainfleet
Jan 10 at 1:03
$begingroup$
Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything.
$endgroup$
– Claude Leibovici
Jan 10 at 7:10
$begingroup$
Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything.
$endgroup$
– Claude Leibovici
Jan 10 at 7:10
$begingroup$
Thanks for catching that, it was a typo, and it should be fixed now.
$endgroup$
– AspiringRoboteer
Jan 10 at 13:29
$begingroup$
Thanks for catching that, it was a typo, and it should be fixed now.
$endgroup$
– AspiringRoboteer
Jan 10 at 13:29
add a comment |
1 Answer
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Within the radius of convergence, one can differentiate a power series term by term. Hence, the derivative is $$frac{d}{dx}sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}=sum_{k=0}^{infty} frac{(-1)^k (3k+1) x^{3k}}{2^{3k}(k!)^3}$$
This is done by using the power rule for derivatives on each term of the power series.
answered Jan 9 at 21:22
greeliousgreelious
465112
$endgroup$
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
add a comment |
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1 Answer
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$begingroup$
Within the radius of convergence, one can differentiate a power series term by term. Hence, the derivative is $$frac{d}{dx}sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}=sum_{k=0}^{infty} frac{(-1)^k (3k+1) x^{3k}}{2^{3k}(k!)^3}$$
This is done by using the power rule for derivatives on each term of the power series.
answered Jan 9 at 21:22
greeliousgreelious
465112
$endgroup$
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
add a comment |
$begingroup$
Within the radius of convergence, one can differentiate a power series term by term. Hence, the derivative is $$frac{d}{dx}sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}=sum_{k=0}^{infty} frac{(-1)^k (3k+1) x^{3k}}{2^{3k}(k!)^3}$$
This is done by using the power rule for derivatives on each term of the power series.
answered Jan 9 at 21:22
greeliousgreelious
465112
$endgroup$
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
add a comment |
$begingroup$
Within the radius of convergence, one can differentiate a power series term by term. Hence, the derivative is $$frac{d}{dx}sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}=sum_{k=0}^{infty} frac{(-1)^k (3k+1) x^{3k}}{2^{3k}(k!)^3}$$
This is done by using the power rule for derivatives on each term of the power series.
answered Jan 9 at 21:22
greeliousgreelious
465112
$endgroup$
Within the radius of convergence, one can differentiate a power series term by term. Hence, the derivative is $$frac{d}{dx}sum_{k=0}^{infty} frac{(-1)^k x^{(3k+1)}}{2^{3k}(k!)^3}=sum_{k=0}^{infty} frac{(-1)^k (3k+1) x^{3k}}{2^{3k}(k!)^3}$$
This is done by using the power rule for derivatives on each term of the power series.
answered Jan 9 at 21:22
greeliousgreelious
465112
edited Jan 26 at 23:40
answered Jan 9 at 21:22
greeliousgreelious
465112
answered Jan 9 at 21:22
greeliousgreelious
465112
answered Jan 9 at 21:22
greeliousgreelious
465112
465112
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
add a comment |
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
$begingroup$
Thank you! I finally understand how to find this!
$endgroup$
– AspiringRoboteer
Jan 9 at 21:30
add a comment |
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$begingroup$
Assuming that $(n!)^3$ is a constant, the summation is $frac {x}{(n!)^3}sum_{k=0}^{infty} (-x^3/2^3)^k,$ which for $|x|<2$ is $frac {x}{(n!)^3}frac {1}{1+x^3/2^3}$.
$endgroup$
– DanielWainfleet
Jan 10 at 1:03
$begingroup$
Taking into account what you wrote, I suppose a typo : $n$ should be $k$ but this does not change anything.
$endgroup$
– Claude Leibovici
Jan 10 at 7:10
$begingroup$
Thanks for catching that, it was a typo, and it should be fixed now.
$endgroup$
– AspiringRoboteer
Jan 10 at 13:29