Write power series as rational function
I need to write the power series:
$sum_{n=1}^infty frac{1}{(x-3)^{2n-1}} - frac{1}{(x-2)^{2n-1}}$
I need to write it as a rational function. I am not sure how to go about doing this.
power-series
asked Dec 7 at 22:39
Elijah
626
add a comment |
I need to write the power series:
$sum_{n=1}^infty frac{1}{(x-3)^{2n-1}} - frac{1}{(x-2)^{2n-1}}$
I need to write it as a rational function. I am not sure how to go about doing this.
power-series
asked Dec 7 at 22:39
Elijah
626
add a comment |
I need to write the power series:
$sum_{n=1}^infty frac{1}{(x-3)^{2n-1}} - frac{1}{(x-2)^{2n-1}}$
I need to write it as a rational function. I am not sure how to go about doing this.
power-series
asked Dec 7 at 22:39
Elijah
626
I need to write the power series:
$sum_{n=1}^infty frac{1}{(x-3)^{2n-1}} - frac{1}{(x-2)^{2n-1}}$
I need to write it as a rational function. I am not sure how to go about doing this.
power-series
power-series
asked Dec 7 at 22:39
Elijah
626
asked Dec 7 at 22:39
Elijah
626
asked Dec 7 at 22:39
Elijah
626
asked Dec 7 at 22:39
Elijah
626
asked Dec 7 at 22:39
Elijah
626
626
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
In general $sum_{n=1}^inftyfrac{1}{u^{2n-1}}=frac{u}{u^2-1}$. Your expression becomes $frac{x-3}{(x-3)^2-1}-frac{x-2}{(x-2)^2-1}$$=frac{(x-3)((x-2)^2-1)-(x-2)((x-3)^2-1)}{((x-3)^2-1)((x-2)^2-1)}$
You could simplify it. Note that $|x-3|gt 1$ and $|x-2| gt 1$ required.
answered Dec 7 at 23:01
herb steinberg
2,4832310
add a comment |
Hint:
This series has the form
$$sum_{n=1}^inftyfrac1{u^{2n-1}}-sum_{n=1}^inftyfrac1{v^{2n-1}}=frac1usum_{n=0}^inftyfrac1{(u^2)^n}-frac1vsum_{n=0}^inftyfrac1{(v^2)^n}.$$
Can you take it from here?
$$
answered Dec 7 at 22:51
Bernard
118k638112
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
In general $sum_{n=1}^inftyfrac{1}{u^{2n-1}}=frac{u}{u^2-1}$. Your expression becomes $frac{x-3}{(x-3)^2-1}-frac{x-2}{(x-2)^2-1}$$=frac{(x-3)((x-2)^2-1)-(x-2)((x-3)^2-1)}{((x-3)^2-1)((x-2)^2-1)}$
You could simplify it. Note that $|x-3|gt 1$ and $|x-2| gt 1$ required.
answered Dec 7 at 23:01
herb steinberg
2,4832310
add a comment |
In general $sum_{n=1}^inftyfrac{1}{u^{2n-1}}=frac{u}{u^2-1}$. Your expression becomes $frac{x-3}{(x-3)^2-1}-frac{x-2}{(x-2)^2-1}$$=frac{(x-3)((x-2)^2-1)-(x-2)((x-3)^2-1)}{((x-3)^2-1)((x-2)^2-1)}$
You could simplify it. Note that $|x-3|gt 1$ and $|x-2| gt 1$ required.
answered Dec 7 at 23:01
herb steinberg
2,4832310
add a comment |
In general $sum_{n=1}^inftyfrac{1}{u^{2n-1}}=frac{u}{u^2-1}$. Your expression becomes $frac{x-3}{(x-3)^2-1}-frac{x-2}{(x-2)^2-1}$$=frac{(x-3)((x-2)^2-1)-(x-2)((x-3)^2-1)}{((x-3)^2-1)((x-2)^2-1)}$
You could simplify it. Note that $|x-3|gt 1$ and $|x-2| gt 1$ required.
answered Dec 7 at 23:01
herb steinberg
2,4832310
In general $sum_{n=1}^inftyfrac{1}{u^{2n-1}}=frac{u}{u^2-1}$. Your expression becomes $frac{x-3}{(x-3)^2-1}-frac{x-2}{(x-2)^2-1}$$=frac{(x-3)((x-2)^2-1)-(x-2)((x-3)^2-1)}{((x-3)^2-1)((x-2)^2-1)}$
You could simplify it. Note that $|x-3|gt 1$ and $|x-2| gt 1$ required.
answered Dec 7 at 23:01
herb steinberg
2,4832310
answered Dec 7 at 23:01
herb steinberg
2,4832310
answered Dec 7 at 23:01
herb steinberg
2,4832310
answered Dec 7 at 23:01
herb steinberg
2,4832310
2,4832310
add a comment |
add a comment |
Hint:
This series has the form
$$sum_{n=1}^inftyfrac1{u^{2n-1}}-sum_{n=1}^inftyfrac1{v^{2n-1}}=frac1usum_{n=0}^inftyfrac1{(u^2)^n}-frac1vsum_{n=0}^inftyfrac1{(v^2)^n}.$$
Can you take it from here?
$$
answered Dec 7 at 22:51
Bernard
118k638112
add a comment |
Hint:
This series has the form
$$sum_{n=1}^inftyfrac1{u^{2n-1}}-sum_{n=1}^inftyfrac1{v^{2n-1}}=frac1usum_{n=0}^inftyfrac1{(u^2)^n}-frac1vsum_{n=0}^inftyfrac1{(v^2)^n}.$$
Can you take it from here?
$$
answered Dec 7 at 22:51
Bernard
118k638112
add a comment |
Hint:
This series has the form
$$sum_{n=1}^inftyfrac1{u^{2n-1}}-sum_{n=1}^inftyfrac1{v^{2n-1}}=frac1usum_{n=0}^inftyfrac1{(u^2)^n}-frac1vsum_{n=0}^inftyfrac1{(v^2)^n}.$$
Can you take it from here?
$$
answered Dec 7 at 22:51
Bernard
118k638112
Hint:
This series has the form
$$sum_{n=1}^inftyfrac1{u^{2n-1}}-sum_{n=1}^inftyfrac1{v^{2n-1}}=frac1usum_{n=0}^inftyfrac1{(u^2)^n}-frac1vsum_{n=0}^inftyfrac1{(v^2)^n}.$$
Can you take it from here?
$$
answered Dec 7 at 22:51
Bernard
118k638112
answered Dec 7 at 22:51
Bernard
118k638112
answered Dec 7 at 22:51
Bernard
118k638112
answered Dec 7 at 22:51
Bernard
118k638112
118k638112
add a comment |
add a comment |
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