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How to find the sum explicitly of this series? Or rather go from this Taylor expansion to a function that represents the sum at a given x?
$sum _n=0^infty :left(n^2-2nright)x^n$
sequences-and-series taylor-expansion
asked 2 hours ago
Alexandre Filho
254
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Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |Â
up vote
2
down vote
favorite
How to find the sum explicitly of this series? Or rather go from this Taylor expansion to a function that represents the sum at a given x?
$sum _n=0^infty :left(n^2-2nright)x^n$
sequences-and-series taylor-expansion
asked 2 hours ago
Alexandre Filho
254
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Notice that $n^2-2n = n(n-2)$. From there, are you familiar with Taylor expansions ? Do you know how to find its antiderivative on the open disk of convergence ?
– Harmonic Sun
2 hours ago
I would like to think that I am familiar, but apparently not. I also don't see how integrating would help. Yes, $x^n$ would eventually become $x^n-1$ and $x^n-2$ ( I think), but I would also have to integrate the $nleft(n-2right)$, wouldn't I?
– Alexandre Filho
1 hour ago
Instead of intergrating, it would be easier to do the other way around : working from the Taylor expansion $sumlimits_n=0^inftyx^n$, and then differentiate. I'm writing an answer right know.
– Harmonic Sun
1 hour ago
Consider the operator "take derivative and then multiply by $x$"
– Hagen von Eitzen
1 hour ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
How to find the sum explicitly of this series? Or rather go from this Taylor expansion to a function that represents the sum at a given x?
$sum _n=0^infty :left(n^2-2nright)x^n$
sequences-and-series taylor-expansion
asked 2 hours ago
Alexandre Filho
254
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
How to find the sum explicitly of this series? Or rather go from this Taylor expansion to a function that represents the sum at a given x?
$sum _n=0^infty :left(n^2-2nright)x^n$
sequences-and-series taylor-expansion
sequences-and-series taylor-expansion
asked 2 hours ago
Alexandre Filho
254
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Alexandre Filho
254
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 1 hour ago
asked 2 hours ago
Alexandre Filho
254
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 hours ago
Alexandre Filho
254
asked 2 hours ago
Alexandre Filho
254
254
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Alexandre Filho is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Notice that $n^2-2n = n(n-2)$. From there, are you familiar with Taylor expansions ? Do you know how to find its antiderivative on the open disk of convergence ?
– Harmonic Sun
2 hours ago
I would like to think that I am familiar, but apparently not. I also don't see how integrating would help. Yes, $x^n$ would eventually become $x^n-1$ and $x^n-2$ ( I think), but I would also have to integrate the $nleft(n-2right)$, wouldn't I?
– Alexandre Filho
1 hour ago
Instead of intergrating, it would be easier to do the other way around : working from the Taylor expansion $sumlimits_n=0^inftyx^n$, and then differentiate. I'm writing an answer right know.
– Harmonic Sun
1 hour ago
Consider the operator "take derivative and then multiply by $x$"
– Hagen von Eitzen
1 hour ago
add a comment |Â
Notice that $n^2-2n = n(n-2)$. From there, are you familiar with Taylor expansions ? Do you know how to find its antiderivative on the open disk of convergence ?
– Harmonic Sun
2 hours ago
I would like to think that I am familiar, but apparently not. I also don't see how integrating would help. Yes, $x^n$ would eventually become $x^n-1$ and $x^n-2$ ( I think), but I would also have to integrate the $nleft(n-2right)$, wouldn't I?
– Alexandre Filho
1 hour ago
Instead of intergrating, it would be easier to do the other way around : working from the Taylor expansion $sumlimits_n=0^inftyx^n$, and then differentiate. I'm writing an answer right know.
– Harmonic Sun
1 hour ago
Consider the operator "take derivative and then multiply by $x$"
– Hagen von Eitzen
1 hour ago
Notice that $n^2-2n = n(n-2)$. From there, are you familiar with Taylor expansions ? Do you know how to find its antiderivative on the open disk of convergence ?
– Harmonic Sun
2 hours ago
Notice that $n^2-2n = n(n-2)$. From there, are you familiar with Taylor expansions ? Do you know how to find its antiderivative on the open disk of convergence ?
– Harmonic Sun
2 hours ago
I would like to think that I am familiar, but apparently not. I also don't see how integrating would help. Yes, $x^n$ would eventually become $x^n-1$ and $x^n-2$ ( I think), but I would also have to integrate the $nleft(n-2right)$, wouldn't I?
– Alexandre Filho
1 hour ago
I would like to think that I am familiar, but apparently not. I also don't see how integrating would help. Yes, $x^n$ would eventually become $x^n-1$ and $x^n-2$ ( I think), but I would also have to integrate the $nleft(n-2right)$, wouldn't I?
– Alexandre Filho
1 hour ago
Instead of intergrating, it would be easier to do the other way around : working from the Taylor expansion $sumlimits_n=0^inftyx^n$, and then differentiate. I'm writing an answer right know.
– Harmonic Sun
1 hour ago
Instead of intergrating, it would be easier to do the other way around : working from the Taylor expansion $sumlimits_n=0^inftyx^n$, and then differentiate. I'm writing an answer right know.
– Harmonic Sun
1 hour ago
Consider the operator "take derivative and then multiply by $x$"
– Hagen von Eitzen
1 hour ago
Consider the operator "take derivative and then multiply by $x$"
– Hagen von Eitzen
1 hour ago
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
5
down vote
accepted
Use $f(x)=sum_n=0^infty x^n=frac11-x$.
We get:
$$f'(x)=sum_n=1^infty nx^n-1=frac1(1-x)^2;\
xf'(x)=sum_n=1^infty nx^n=fracx(1-x)^2 Rightarrow colorredsum_n=0^infty nx^n=fracx(1-x)^2;\
[xf'(x)]'=f'(x)+xf''(x)=sum_n=1^infty n^2x^n-1=frac1+x(1-x)^3;\
xf'(x)+x^2f''(x)=sum_n=1^infty n^2x^n=fracx(1+x)(1-x)^3 Rightarrow colorbluesum_n=0^infty n^2x^n=fracx(1+x)(1-x)^3;\
==============================================\
sum _n=0^infty :left(n^2-2nright)x^n=colorbluesum_n=0^infty n^2x^n-2colorredsum_n=0^inftynx^n=colorbluefracx(1+x)(1-x)^3-2colorredfracx(1-x)^2=\
fracx+x^2-2x(1-x)(1-x)^3=frac3x^2-x(1-x)^3.$$
answered 1 hour ago
farruhota
15.9k2734
add a comment |Â
up vote
2
down vote
This series can also be evaluated in the following way:
beginalign
sum_n=0^infty (n^2 - 2n) , x^n &= sum_n=0^infty n (n-2) , x^n \
&= sum_n=0^infty (n(n-1) - n) x^n \
&= x^2 , sum_n=0^infty n(n-1) , x^n-2 - x , sum_n=0^infty n , x^n-1 \
&= x^2 , fracd^2dx^2 , sum_n=0^infty x^n - x , fracddx , sum_n=0^infty x^n \
&= x^2 , fracd^2dx^2 , frac11-x - x , fracddx frac11-x \
&= fracx(3x -1)(1-x)^3
endalign
answered 1 hour ago
Leucippus
19.2k102869
add a comment |Â
up vote
0
down vote
One powerful way to approach these problems is to think of the sum as a function of $x$, meaning. but first let's let's separate the sum into two parts for reasons that will become clear later:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - x sum_n=0^infty n x^n-1$$
For the first summation we have:
$$ sum_n=0^infty (n^2 - n) x^n-2
= sum_n=0^infty fracddx n x^n-1$$
$$ = fracddx sum_n=0^infty fracddx x^n
= fracd^2dx^2 sum_n=0 x^n$$
$$=fracd^2dx^2 frac11-x = frac2(1-x)^3$$.
Similarly for the second summation we can derive:
$$sum_n=0^infty n x^n-1 = sum_n=0^infty fracddx x^n$$
$$ = fracddx sum_n=0^infty x^n = fracddx frac11-x = frac1(1-x)^2$$
Now you can see the reason why the terms were separated that way. It was for the coefficients to be cancelled out after integration. Now back to the original summation we have:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - xsum_n=0^infty n x^n-1$$
$$ = x^2 frac2(1-x)^3 - xfrac1(1-x)^2 $$
$$= frac2x^2 - (1-x)x(1-x)^3 = fracx(3x-1)(1-x)^3$$.
There are a couple of points to be mentioned. First is that the formula above is only correct for values $|x|<1$ as clearly the series will be divergent for other values. The second point is that while integration and sum over finite number of functions can be interchanged it's far from trivial if the same holds for infinite summation as in this question, which one should be careful about.
answered 1 hour ago
kvphxga
1163
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Use $f(x)=sum_n=0^infty x^n=frac11-x$.
We get:
$$f'(x)=sum_n=1^infty nx^n-1=frac1(1-x)^2;\
xf'(x)=sum_n=1^infty nx^n=fracx(1-x)^2 Rightarrow colorredsum_n=0^infty nx^n=fracx(1-x)^2;\
[xf'(x)]'=f'(x)+xf''(x)=sum_n=1^infty n^2x^n-1=frac1+x(1-x)^3;\
xf'(x)+x^2f''(x)=sum_n=1^infty n^2x^n=fracx(1+x)(1-x)^3 Rightarrow colorbluesum_n=0^infty n^2x^n=fracx(1+x)(1-x)^3;\
==============================================\
sum _n=0^infty :left(n^2-2nright)x^n=colorbluesum_n=0^infty n^2x^n-2colorredsum_n=0^inftynx^n=colorbluefracx(1+x)(1-x)^3-2colorredfracx(1-x)^2=\
fracx+x^2-2x(1-x)(1-x)^3=frac3x^2-x(1-x)^3.$$
answered 1 hour ago
farruhota
15.9k2734
add a comment |Â
up vote
5
down vote
accepted
Use $f(x)=sum_n=0^infty x^n=frac11-x$.
We get:
$$f'(x)=sum_n=1^infty nx^n-1=frac1(1-x)^2;\
xf'(x)=sum_n=1^infty nx^n=fracx(1-x)^2 Rightarrow colorredsum_n=0^infty nx^n=fracx(1-x)^2;\
[xf'(x)]'=f'(x)+xf''(x)=sum_n=1^infty n^2x^n-1=frac1+x(1-x)^3;\
xf'(x)+x^2f''(x)=sum_n=1^infty n^2x^n=fracx(1+x)(1-x)^3 Rightarrow colorbluesum_n=0^infty n^2x^n=fracx(1+x)(1-x)^3;\
==============================================\
sum _n=0^infty :left(n^2-2nright)x^n=colorbluesum_n=0^infty n^2x^n-2colorredsum_n=0^inftynx^n=colorbluefracx(1+x)(1-x)^3-2colorredfracx(1-x)^2=\
fracx+x^2-2x(1-x)(1-x)^3=frac3x^2-x(1-x)^3.$$
answered 1 hour ago
farruhota
15.9k2734
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Use $f(x)=sum_n=0^infty x^n=frac11-x$.
We get:
$$f'(x)=sum_n=1^infty nx^n-1=frac1(1-x)^2;\
xf'(x)=sum_n=1^infty nx^n=fracx(1-x)^2 Rightarrow colorredsum_n=0^infty nx^n=fracx(1-x)^2;\
[xf'(x)]'=f'(x)+xf''(x)=sum_n=1^infty n^2x^n-1=frac1+x(1-x)^3;\
xf'(x)+x^2f''(x)=sum_n=1^infty n^2x^n=fracx(1+x)(1-x)^3 Rightarrow colorbluesum_n=0^infty n^2x^n=fracx(1+x)(1-x)^3;\
==============================================\
sum _n=0^infty :left(n^2-2nright)x^n=colorbluesum_n=0^infty n^2x^n-2colorredsum_n=0^inftynx^n=colorbluefracx(1+x)(1-x)^3-2colorredfracx(1-x)^2=\
fracx+x^2-2x(1-x)(1-x)^3=frac3x^2-x(1-x)^3.$$
answered 1 hour ago
farruhota
15.9k2734
Use $f(x)=sum_n=0^infty x^n=frac11-x$.
We get:
$$f'(x)=sum_n=1^infty nx^n-1=frac1(1-x)^2;\
xf'(x)=sum_n=1^infty nx^n=fracx(1-x)^2 Rightarrow colorredsum_n=0^infty nx^n=fracx(1-x)^2;\
[xf'(x)]'=f'(x)+xf''(x)=sum_n=1^infty n^2x^n-1=frac1+x(1-x)^3;\
xf'(x)+x^2f''(x)=sum_n=1^infty n^2x^n=fracx(1+x)(1-x)^3 Rightarrow colorbluesum_n=0^infty n^2x^n=fracx(1+x)(1-x)^3;\
==============================================\
sum _n=0^infty :left(n^2-2nright)x^n=colorbluesum_n=0^infty n^2x^n-2colorredsum_n=0^inftynx^n=colorbluefracx(1+x)(1-x)^3-2colorredfracx(1-x)^2=\
fracx+x^2-2x(1-x)(1-x)^3=frac3x^2-x(1-x)^3.$$
answered 1 hour ago
farruhota
15.9k2734
answered 1 hour ago
farruhota
15.9k2734
answered 1 hour ago
farruhota
15.9k2734
answered 1 hour ago
farruhota
15.9k2734
15.9k2734
add a comment |Â
add a comment |Â
up vote
2
down vote
This series can also be evaluated in the following way:
beginalign
sum_n=0^infty (n^2 - 2n) , x^n &= sum_n=0^infty n (n-2) , x^n \
&= sum_n=0^infty (n(n-1) - n) x^n \
&= x^2 , sum_n=0^infty n(n-1) , x^n-2 - x , sum_n=0^infty n , x^n-1 \
&= x^2 , fracd^2dx^2 , sum_n=0^infty x^n - x , fracddx , sum_n=0^infty x^n \
&= x^2 , fracd^2dx^2 , frac11-x - x , fracddx frac11-x \
&= fracx(3x -1)(1-x)^3
endalign
answered 1 hour ago
Leucippus
19.2k102869
add a comment |Â
up vote
2
down vote
This series can also be evaluated in the following way:
beginalign
sum_n=0^infty (n^2 - 2n) , x^n &= sum_n=0^infty n (n-2) , x^n \
&= sum_n=0^infty (n(n-1) - n) x^n \
&= x^2 , sum_n=0^infty n(n-1) , x^n-2 - x , sum_n=0^infty n , x^n-1 \
&= x^2 , fracd^2dx^2 , sum_n=0^infty x^n - x , fracddx , sum_n=0^infty x^n \
&= x^2 , fracd^2dx^2 , frac11-x - x , fracddx frac11-x \
&= fracx(3x -1)(1-x)^3
endalign
answered 1 hour ago
Leucippus
19.2k102869
add a comment |Â
up vote
2
down vote
up vote
2
down vote
This series can also be evaluated in the following way:
beginalign
sum_n=0^infty (n^2 - 2n) , x^n &= sum_n=0^infty n (n-2) , x^n \
&= sum_n=0^infty (n(n-1) - n) x^n \
&= x^2 , sum_n=0^infty n(n-1) , x^n-2 - x , sum_n=0^infty n , x^n-1 \
&= x^2 , fracd^2dx^2 , sum_n=0^infty x^n - x , fracddx , sum_n=0^infty x^n \
&= x^2 , fracd^2dx^2 , frac11-x - x , fracddx frac11-x \
&= fracx(3x -1)(1-x)^3
endalign
answered 1 hour ago
Leucippus
19.2k102869
This series can also be evaluated in the following way:
beginalign
sum_n=0^infty (n^2 - 2n) , x^n &= sum_n=0^infty n (n-2) , x^n \
&= sum_n=0^infty (n(n-1) - n) x^n \
&= x^2 , sum_n=0^infty n(n-1) , x^n-2 - x , sum_n=0^infty n , x^n-1 \
&= x^2 , fracd^2dx^2 , sum_n=0^infty x^n - x , fracddx , sum_n=0^infty x^n \
&= x^2 , fracd^2dx^2 , frac11-x - x , fracddx frac11-x \
&= fracx(3x -1)(1-x)^3
endalign
answered 1 hour ago
Leucippus
19.2k102869
answered 1 hour ago
Leucippus
19.2k102869
answered 1 hour ago
Leucippus
19.2k102869
answered 1 hour ago
Leucippus
19.2k102869
19.2k102869
add a comment |Â
add a comment |Â
up vote
0
down vote
One powerful way to approach these problems is to think of the sum as a function of $x$, meaning. but first let's let's separate the sum into two parts for reasons that will become clear later:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - x sum_n=0^infty n x^n-1$$
For the first summation we have:
$$ sum_n=0^infty (n^2 - n) x^n-2
= sum_n=0^infty fracddx n x^n-1$$
$$ = fracddx sum_n=0^infty fracddx x^n
= fracd^2dx^2 sum_n=0 x^n$$
$$=fracd^2dx^2 frac11-x = frac2(1-x)^3$$.
Similarly for the second summation we can derive:
$$sum_n=0^infty n x^n-1 = sum_n=0^infty fracddx x^n$$
$$ = fracddx sum_n=0^infty x^n = fracddx frac11-x = frac1(1-x)^2$$
Now you can see the reason why the terms were separated that way. It was for the coefficients to be cancelled out after integration. Now back to the original summation we have:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - xsum_n=0^infty n x^n-1$$
$$ = x^2 frac2(1-x)^3 - xfrac1(1-x)^2 $$
$$= frac2x^2 - (1-x)x(1-x)^3 = fracx(3x-1)(1-x)^3$$.
There are a couple of points to be mentioned. First is that the formula above is only correct for values $|x|<1$ as clearly the series will be divergent for other values. The second point is that while integration and sum over finite number of functions can be interchanged it's far from trivial if the same holds for infinite summation as in this question, which one should be careful about.
answered 1 hour ago
kvphxga
1163
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
add a comment |Â
up vote
0
down vote
One powerful way to approach these problems is to think of the sum as a function of $x$, meaning. but first let's let's separate the sum into two parts for reasons that will become clear later:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - x sum_n=0^infty n x^n-1$$
For the first summation we have:
$$ sum_n=0^infty (n^2 - n) x^n-2
= sum_n=0^infty fracddx n x^n-1$$
$$ = fracddx sum_n=0^infty fracddx x^n
= fracd^2dx^2 sum_n=0 x^n$$
$$=fracd^2dx^2 frac11-x = frac2(1-x)^3$$.
Similarly for the second summation we can derive:
$$sum_n=0^infty n x^n-1 = sum_n=0^infty fracddx x^n$$
$$ = fracddx sum_n=0^infty x^n = fracddx frac11-x = frac1(1-x)^2$$
Now you can see the reason why the terms were separated that way. It was for the coefficients to be cancelled out after integration. Now back to the original summation we have:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - xsum_n=0^infty n x^n-1$$
$$ = x^2 frac2(1-x)^3 - xfrac1(1-x)^2 $$
$$= frac2x^2 - (1-x)x(1-x)^3 = fracx(3x-1)(1-x)^3$$.
There are a couple of points to be mentioned. First is that the formula above is only correct for values $|x|<1$ as clearly the series will be divergent for other values. The second point is that while integration and sum over finite number of functions can be interchanged it's far from trivial if the same holds for infinite summation as in this question, which one should be careful about.
answered 1 hour ago
kvphxga
1163
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
add a comment |Â
up vote
0
down vote
up vote
0
down vote
One powerful way to approach these problems is to think of the sum as a function of $x$, meaning. but first let's let's separate the sum into two parts for reasons that will become clear later:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - x sum_n=0^infty n x^n-1$$
For the first summation we have:
$$ sum_n=0^infty (n^2 - n) x^n-2
= sum_n=0^infty fracddx n x^n-1$$
$$ = fracddx sum_n=0^infty fracddx x^n
= fracd^2dx^2 sum_n=0 x^n$$
$$=fracd^2dx^2 frac11-x = frac2(1-x)^3$$.
Similarly for the second summation we can derive:
$$sum_n=0^infty n x^n-1 = sum_n=0^infty fracddx x^n$$
$$ = fracddx sum_n=0^infty x^n = fracddx frac11-x = frac1(1-x)^2$$
Now you can see the reason why the terms were separated that way. It was for the coefficients to be cancelled out after integration. Now back to the original summation we have:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - xsum_n=0^infty n x^n-1$$
$$ = x^2 frac2(1-x)^3 - xfrac1(1-x)^2 $$
$$= frac2x^2 - (1-x)x(1-x)^3 = fracx(3x-1)(1-x)^3$$.
There are a couple of points to be mentioned. First is that the formula above is only correct for values $|x|<1$ as clearly the series will be divergent for other values. The second point is that while integration and sum over finite number of functions can be interchanged it's far from trivial if the same holds for infinite summation as in this question, which one should be careful about.
answered 1 hour ago
kvphxga
1163
One powerful way to approach these problems is to think of the sum as a function of $x$, meaning. but first let's let's separate the sum into two parts for reasons that will become clear later:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - x sum_n=0^infty n x^n-1$$
For the first summation we have:
$$ sum_n=0^infty (n^2 - n) x^n-2
= sum_n=0^infty fracddx n x^n-1$$
$$ = fracddx sum_n=0^infty fracddx x^n
= fracd^2dx^2 sum_n=0 x^n$$
$$=fracd^2dx^2 frac11-x = frac2(1-x)^3$$.
Similarly for the second summation we can derive:
$$sum_n=0^infty n x^n-1 = sum_n=0^infty fracddx x^n$$
$$ = fracddx sum_n=0^infty x^n = fracddx frac11-x = frac1(1-x)^2$$
Now you can see the reason why the terms were separated that way. It was for the coefficients to be cancelled out after integration. Now back to the original summation we have:
$$sum_n=0^infty (n^2-2n)x^n = x^2 sum_n=0^infty (n^2 -n)x^n-2 - xsum_n=0^infty n x^n-1$$
$$ = x^2 frac2(1-x)^3 - xfrac1(1-x)^2 $$
$$= frac2x^2 - (1-x)x(1-x)^3 = fracx(3x-1)(1-x)^3$$.
There are a couple of points to be mentioned. First is that the formula above is only correct for values $|x|<1$ as clearly the series will be divergent for other values. The second point is that while integration and sum over finite number of functions can be interchanged it's far from trivial if the same holds for infinite summation as in this question, which one should be careful about.
answered 1 hour ago
kvphxga
1163
edited 1 min ago
answered 1 hour ago
kvphxga
1163
answered 1 hour ago
kvphxga
1163
answered 1 hour ago
kvphxga
1163
1163
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
add a comment |Â
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
I used Wolfram to validate the answers and your answer. However, only when multiplied by x $frac3-x(1-x)^3$ leads to the right result. I tested for x = 1/2, which the sum should be 2
– Alexandre Filho
1 hour ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
Yes, I missed the extra $x^2$ and $x$ in calculating the final answer. Now it's consistent with others' results.
– kvphxga
13 secs ago
add a comment |Â
Alexandre Filho is a new contributor. Be nice, and check out our Code of Conduct.
Alexandre Filho is a new contributor. Be nice, and check out our Code of Conduct.
Alexandre Filho is a new contributor. Be nice, and check out our Code of Conduct.
Alexandre Filho is a new contributor. Be nice, and check out our Code of Conduct.
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Notice that $n^2-2n = n(n-2)$. From there, are you familiar with Taylor expansions ? Do you know how to find its antiderivative on the open disk of convergence ?
– Harmonic Sun
2 hours ago
I would like to think that I am familiar, but apparently not. I also don't see how integrating would help. Yes, $x^n$ would eventually become $x^n-1$ and $x^n-2$ ( I think), but I would also have to integrate the $nleft(n-2right)$, wouldn't I?
– Alexandre Filho
1 hour ago
Instead of intergrating, it would be easier to do the other way around : working from the Taylor expansion $sumlimits_n=0^inftyx^n$, and then differentiate. I'm writing an answer right know.
– Harmonic Sun
1 hour ago
Consider the operator "take derivative and then multiply by $x$"
– Hagen von Eitzen
1 hour ago