Series representation of $1/x^2$?
$begingroup$
According to WolframAlpha the series representation of $1/x^2$ is
$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$
Can somebody tell me how to prove this result?
sequences-and-series
edited Dec 19 '18 at 15:14
amWhy
1
asked Dec 19 '18 at 15:09
LightLight
111
$endgroup$
add a comment |
$begingroup$
According to WolframAlpha the series representation of $1/x^2$ is
$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$
Can somebody tell me how to prove this result?
sequences-and-series
edited Dec 19 '18 at 15:14
amWhy
1
asked Dec 19 '18 at 15:09
LightLight
111
$endgroup$
$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19
add a comment |
$begingroup$
According to WolframAlpha the series representation of $1/x^2$ is
$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$
Can somebody tell me how to prove this result?
sequences-and-series
edited Dec 19 '18 at 15:14
amWhy
1
asked Dec 19 '18 at 15:09
LightLight
111
$endgroup$
According to WolframAlpha the series representation of $1/x^2$ is
$$frac{1}{x^2} = sum_{n=0}^{infty} (x-1)^{n} (-1)^n (n+1) $$
Can somebody tell me how to prove this result?
sequences-and-series
sequences-and-series
edited Dec 19 '18 at 15:14
amWhy
1
asked Dec 19 '18 at 15:09
LightLight
111
edited Dec 19 '18 at 15:14
amWhy
1
asked Dec 19 '18 at 15:09
LightLight
111
edited Dec 19 '18 at 15:14
amWhy
1
edited Dec 19 '18 at 15:14
amWhy
1
edited Dec 19 '18 at 15:14
amWhy
1
1
asked Dec 19 '18 at 15:09
LightLight
111
asked Dec 19 '18 at 15:09
LightLight
111
asked Dec 19 '18 at 15:09
LightLight
111
111
$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19
add a comment |
$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19
$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19
$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19
add a comment |
4 Answers
4
active
oldest
votes
$begingroup$
This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.
Alternatively, you can start from the geometric series
$$
frac{1}{1-u} = 1 + u + u^2 + cdots,
$$
differentiate term by term with respect to $u$ and substitute $u =1-x$.
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
add a comment |
$begingroup$
This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
$$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
which is valid for $|y|<1$. Take the geometric series
$$frac1{1-y}=sum_{n=0}^infty y^n$$
and either differentiate it or square it.
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
Hint: start from the geometric series
$$sum_{n=0}^infty x^n = frac1{1-x}.$$
Alternatively: Taylor at $x = 1$.
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
$endgroup$
add a comment |
$begingroup$
By the geometric series, you obtain
$$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$
You can also consider the Taylor Series, which gets
$$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$
and see what happens at $a = 1$...
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
$endgroup$
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.
Alternatively, you can start from the geometric series
$$
frac{1}{1-u} = 1 + u + u^2 + cdots,
$$
differentiate term by term with respect to $u$ and substitute $u =1-x$.
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
add a comment |
$begingroup$
This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.
Alternatively, you can start from the geometric series
$$
frac{1}{1-u} = 1 + u + u^2 + cdots,
$$
differentiate term by term with respect to $u$ and substitute $u =1-x$.
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
add a comment |
$begingroup$
This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.
Alternatively, you can start from the geometric series
$$
frac{1}{1-u} = 1 + u + u^2 + cdots,
$$
differentiate term by term with respect to $u$ and substitute $u =1-x$.
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
$endgroup$
This is just the ordinary Taylor series for $f(x) = 1/x^2$ around the point $x=1$. So you find the coefficients by finding the derivatives there.
Alternatively, you can start from the geometric series
$$
frac{1}{1-u} = 1 + u + u^2 + cdots,
$$
differentiate term by term with respect to $u$ and substitute $u =1-x$.
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
answered Dec 19 '18 at 15:15
Ethan BolkerEthan Bolker
43.4k551116
43.4k551116
add a comment |
add a comment |
$begingroup$
This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
$$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
which is valid for $|y|<1$. Take the geometric series
$$frac1{1-y}=sum_{n=0}^infty y^n$$
and either differentiate it or square it.
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
$$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
which is valid for $|y|<1$. Take the geometric series
$$frac1{1-y}=sum_{n=0}^infty y^n$$
and either differentiate it or square it.
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
add a comment |
$begingroup$
This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
$$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
which is valid for $|y|<1$. Take the geometric series
$$frac1{1-y}=sum_{n=0}^infty y^n$$
and either differentiate it or square it.
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
$endgroup$
This is valid for $|x-1|<1$. Setting $y=1-x$, it's equivalent to
$$frac1{(1-y)^2}=sum_{n=0}^infty (n+1)y^n$$
which is valid for $|y|<1$. Take the geometric series
$$frac1{1-y}=sum_{n=0}^infty y^n$$
and either differentiate it or square it.
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
answered Dec 19 '18 at 15:14
Lord Shark the UnknownLord Shark the Unknown
104k1160132
104k1160132
add a comment |
add a comment |
$begingroup$
Hint: start from the geometric series
$$sum_{n=0}^infty x^n = frac1{1-x}.$$
Alternatively: Taylor at $x = 1$.
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
$endgroup$
add a comment |
$begingroup$
Hint: start from the geometric series
$$sum_{n=0}^infty x^n = frac1{1-x}.$$
Alternatively: Taylor at $x = 1$.
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
$endgroup$
add a comment |
$begingroup$
Hint: start from the geometric series
$$sum_{n=0}^infty x^n = frac1{1-x}.$$
Alternatively: Taylor at $x = 1$.
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
$endgroup$
Hint: start from the geometric series
$$sum_{n=0}^infty x^n = frac1{1-x}.$$
Alternatively: Taylor at $x = 1$.
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
answered Dec 19 '18 at 15:13
Martín-Blas Pérez PinillaMartín-Blas Pérez Pinilla
34.4k42871
34.4k42871
add a comment |
add a comment |
$begingroup$
By the geometric series, you obtain
$$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$
You can also consider the Taylor Series, which gets
$$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$
and see what happens at $a = 1$...
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
By the geometric series, you obtain
$$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$
You can also consider the Taylor Series, which gets
$$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$
and see what happens at $a = 1$...
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
$endgroup$
add a comment |
$begingroup$
By the geometric series, you obtain
$$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$
You can also consider the Taylor Series, which gets
$$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$
and see what happens at $a = 1$...
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
$endgroup$
By the geometric series, you obtain
$$sum_{n geq 0} r^n = frac{1}{1-r}; vert rvert < 1$$
You can also consider the Taylor Series, which gets
$$f(x) = a^{-2}-2a^{-3}(x-a)+frac{6a^{-4}}{2!}(x-a)^2-…$$
and see what happens at $a = 1$...
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
answered Dec 19 '18 at 15:29
KM101KM101
6,0251525
6,0251525
add a comment |
add a comment |
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$begingroup$
This is not the series representation. It is the series representation about the point $x=1$, which is valid for $|x-1|<1$. For any non-zero $ainBbb R$, there is a series representation for $1/x^2$ about the point $x=a$, valid for $|x-a|<a$; and this representation is different for every $a$.
$endgroup$
– TonyK
Dec 19 '18 at 15:19