Find first five non-zero terms of power series
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I have the function f(x) = $dfrac{2x}{left(x-5right)^2}$, and I'm supposed to "find the first five non-zero terms of power series representation centered at x = 0."
Using $frac{f^{(n)}(0)}{n!}$, the first non-zero terms I get are:
C1 = 2/25
C2 = 4/125
C3 = 6/625
C4 = 8/3125
C5 = 2/3125
However, WebWork marks these as incorrect. Any alternative methods or corrections to the work above are gladly appreciated.
calculus sequences-and-series power-series
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
$endgroup$
add a comment |
$begingroup$
I have the function f(x) = $dfrac{2x}{left(x-5right)^2}$, and I'm supposed to "find the first five non-zero terms of power series representation centered at x = 0."
Using $frac{f^{(n)}(0)}{n!}$, the first non-zero terms I get are:
C1 = 2/25
C2 = 4/125
C3 = 6/625
C4 = 8/3125
C5 = 2/3125
However, WebWork marks these as incorrect. Any alternative methods or corrections to the work above are gladly appreciated.
calculus sequences-and-series power-series
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
$endgroup$
$begingroup$
those are the terms wolfram alpha gives.
$endgroup$
– Dando18
Dec 8 '18 at 20:36
$begingroup$
Does that mean the terms I listed are correct?
$endgroup$
– BOBBY SHMURDA
Dec 8 '18 at 21:10
$begingroup$
It seems so... Unless the question is asking for something else.
$endgroup$
– Dando18
Dec 8 '18 at 21:54
add a comment |
$begingroup$
I have the function f(x) = $dfrac{2x}{left(x-5right)^2}$, and I'm supposed to "find the first five non-zero terms of power series representation centered at x = 0."
Using $frac{f^{(n)}(0)}{n!}$, the first non-zero terms I get are:
C1 = 2/25
C2 = 4/125
C3 = 6/625
C4 = 8/3125
C5 = 2/3125
However, WebWork marks these as incorrect. Any alternative methods or corrections to the work above are gladly appreciated.
calculus sequences-and-series power-series
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
$endgroup$
I have the function f(x) = $dfrac{2x}{left(x-5right)^2}$, and I'm supposed to "find the first five non-zero terms of power series representation centered at x = 0."
Using $frac{f^{(n)}(0)}{n!}$, the first non-zero terms I get are:
C1 = 2/25
C2 = 4/125
C3 = 6/625
C4 = 8/3125
C5 = 2/3125
However, WebWork marks these as incorrect. Any alternative methods or corrections to the work above are gladly appreciated.
calculus sequences-and-series power-series
calculus sequences-and-series power-series
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
asked Dec 8 '18 at 20:25
BOBBY SHMURDABOBBY SHMURDA
102
102
$begingroup$
those are the terms wolfram alpha gives.
$endgroup$
– Dando18
Dec 8 '18 at 20:36
$begingroup$
Does that mean the terms I listed are correct?
$endgroup$
– BOBBY SHMURDA
Dec 8 '18 at 21:10
$begingroup$
It seems so... Unless the question is asking for something else.
$endgroup$
– Dando18
Dec 8 '18 at 21:54
add a comment |
$begingroup$
those are the terms wolfram alpha gives.
$endgroup$
– Dando18
Dec 8 '18 at 20:36
$begingroup$
Does that mean the terms I listed are correct?
$endgroup$
– BOBBY SHMURDA
Dec 8 '18 at 21:10
$begingroup$
It seems so... Unless the question is asking for something else.
$endgroup$
– Dando18
Dec 8 '18 at 21:54
$begingroup$
those are the terms wolfram alpha gives.
$endgroup$
– Dando18
Dec 8 '18 at 20:36
$begingroup$
those are the terms wolfram alpha gives.
$endgroup$
– Dando18
Dec 8 '18 at 20:36
$begingroup$
Does that mean the terms I listed are correct?
$endgroup$
– BOBBY SHMURDA
Dec 8 '18 at 21:10
$begingroup$
Does that mean the terms I listed are correct?
$endgroup$
– BOBBY SHMURDA
Dec 8 '18 at 21:10
$begingroup$
It seems so... Unless the question is asking for something else.
$endgroup$
– Dando18
Dec 8 '18 at 21:54
$begingroup$
It seems so... Unless the question is asking for something else.
$endgroup$
– Dando18
Dec 8 '18 at 21:54
add a comment |
1 Answer
1
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$begingroup$
Hint:
You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$frac{2x}{left(x-5right)^2}=frac{2x}{25}frac 1{Bigl(1-cfrac x5Bigr)^2}=frac{2x}5left(frac 1{1-smash[b]{cfrac x5}}right)'$$
and use the power series expansion $;dfrac1{1-u}=1+u+u^2+dotsm$
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
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add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Hint:
You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$frac{2x}{left(x-5right)^2}=frac{2x}{25}frac 1{Bigl(1-cfrac x5Bigr)^2}=frac{2x}5left(frac 1{1-smash[b]{cfrac x5}}right)'$$
and use the power series expansion $;dfrac1{1-u}=1+u+u^2+dotsm$
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Hint:
You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$frac{2x}{left(x-5right)^2}=frac{2x}{25}frac 1{Bigl(1-cfrac x5Bigr)^2}=frac{2x}5left(frac 1{1-smash[b]{cfrac x5}}right)'$$
and use the power series expansion $;dfrac1{1-u}=1+u+u^2+dotsm$
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
$endgroup$
add a comment |
$begingroup$
Hint:
You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$frac{2x}{left(x-5right)^2}=frac{2x}{25}frac 1{Bigl(1-cfrac x5Bigr)^2}=frac{2x}5left(frac 1{1-smash[b]{cfrac x5}}right)'$$
and use the power series expansion $;dfrac1{1-u}=1+u+u^2+dotsm$
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
$endgroup$
Hint:
You don't have to compute the successive derivatives. Instead rewrite the fraction as
$$frac{2x}{left(x-5right)^2}=frac{2x}{25}frac 1{Bigl(1-cfrac x5Bigr)^2}=frac{2x}5left(frac 1{1-smash[b]{cfrac x5}}right)'$$
and use the power series expansion $;dfrac1{1-u}=1+u+u^2+dotsm$
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
answered Dec 8 '18 at 20:58
BernardBernard
119k740113
119k740113
add a comment |
add a comment |
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$begingroup$
those are the terms wolfram alpha gives.
$endgroup$
– Dando18
Dec 8 '18 at 20:36
$begingroup$
Does that mean the terms I listed are correct?
$endgroup$
– BOBBY SHMURDA
Dec 8 '18 at 21:10
$begingroup$
It seems so... Unless the question is asking for something else.
$endgroup$
– Dando18
Dec 8 '18 at 21:54