On the sum of Double Power Series $sumlimits_-infty^+inftyspacefrac2^nx^2^n+1$
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Let $,xinmathbbC,$, show that:
$$ S_small-=sum_n=1^inftyspacefrac2^-nx^2^-n+1,=frac1logx-frac1x-1qquadqquadqquadtag1 $$
$$ S_small+=sum_n=0^inftyspacefrac2^+nx^2^+n+1,=frac1x-1qquadqquadqquadqquadqquadtag2 $$
$$ S,,=sum_-infty^+inftyspacefrac2^n,x^2^n,,,+1,=S_small-+S_small+=frac1logxqquadqquadqquadtag3 $$
And find the convergence range of each sum.
In a similar question, the OP start by introducing $,(2),$ as a known identity, and gives a turly remarkable proof in the comment:
$$ sum_n=0^mlog(x^2^n+y^2^n) =logprod_n=0^m(x^2^n+y^2^n) =logfrac(x-y)x-yprod_n=0^m(x^2^n+y^2^n) =logfracx^2^m+1-y^2^m+1x-y $$
Then differentiate with respect to $y$, substitute $y=1$, and limit $,mtoinfty,$ give the result.
Could this approach work to proof $,(1),$? notice for $x=1$:
$$ S_small-=sum_n=1^inftyfrac12^n+1,=frac12 qquad=qquad lim_small xto1left[frac1logx-frac1x-1right]=frac12 $$
New alternative methods are highly welcomed. Thanks.
sequences-and-series summation power-series
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
add a comment |Â
up vote
4
down vote
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Let $,xinmathbbC,$, show that:
$$ S_small-=sum_n=1^inftyspacefrac2^-nx^2^-n+1,=frac1logx-frac1x-1qquadqquadqquadtag1 $$
$$ S_small+=sum_n=0^inftyspacefrac2^+nx^2^+n+1,=frac1x-1qquadqquadqquadqquadqquadtag2 $$
$$ S,,=sum_-infty^+inftyspacefrac2^n,x^2^n,,,+1,=S_small-+S_small+=frac1logxqquadqquadqquadtag3 $$
And find the convergence range of each sum.
In a similar question, the OP start by introducing $,(2),$ as a known identity, and gives a turly remarkable proof in the comment:
$$ sum_n=0^mlog(x^2^n+y^2^n) =logprod_n=0^m(x^2^n+y^2^n) =logfrac(x-y)x-yprod_n=0^m(x^2^n+y^2^n) =logfracx^2^m+1-y^2^m+1x-y $$
Then differentiate with respect to $y$, substitute $y=1$, and limit $,mtoinfty,$ give the result.
Could this approach work to proof $,(1),$? notice for $x=1$:
$$ S_small-=sum_n=1^inftyfrac12^n+1,=frac12 qquad=qquad lim_small xto1left[frac1logx-frac1x-1right]=frac12 $$
New alternative methods are highly welcomed. Thanks.
sequences-and-series summation power-series
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
You might consider stopping altogether to play with the sizes of fonts, the result being exhausting to read...
– Did
Aug 26 at 21:31
@Did Apologizing.
– Hazem Orabi
Aug 26 at 21:42
1
The more general identity exists: $$sum_n=0^m frac2^n y^2^nx^2^n+y^2^n = fracyx-y-frac2^m+1y^2^m+1x^2^m+1-y^2^m+1 $$ For $y=1$ and $m rightarrow +infty$ we get what you desire.
– Tolaso
Aug 26 at 21:45
@Tolaso Again proofing (2). Thanks.
– Hazem Orabi
Aug 26 at 21:51
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $,xinmathbbC,$, show that:
$$ S_small-=sum_n=1^inftyspacefrac2^-nx^2^-n+1,=frac1logx-frac1x-1qquadqquadqquadtag1 $$
$$ S_small+=sum_n=0^inftyspacefrac2^+nx^2^+n+1,=frac1x-1qquadqquadqquadqquadqquadtag2 $$
$$ S,,=sum_-infty^+inftyspacefrac2^n,x^2^n,,,+1,=S_small-+S_small+=frac1logxqquadqquadqquadtag3 $$
And find the convergence range of each sum.
In a similar question, the OP start by introducing $,(2),$ as a known identity, and gives a turly remarkable proof in the comment:
$$ sum_n=0^mlog(x^2^n+y^2^n) =logprod_n=0^m(x^2^n+y^2^n) =logfrac(x-y)x-yprod_n=0^m(x^2^n+y^2^n) =logfracx^2^m+1-y^2^m+1x-y $$
Then differentiate with respect to $y$, substitute $y=1$, and limit $,mtoinfty,$ give the result.
Could this approach work to proof $,(1),$? notice for $x=1$:
$$ S_small-=sum_n=1^inftyfrac12^n+1,=frac12 qquad=qquad lim_small xto1left[frac1logx-frac1x-1right]=frac12 $$
New alternative methods are highly welcomed. Thanks.
sequences-and-series summation power-series
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
Let $,xinmathbbC,$, show that:
$$ S_small-=sum_n=1^inftyspacefrac2^-nx^2^-n+1,=frac1logx-frac1x-1qquadqquadqquadtag1 $$
$$ S_small+=sum_n=0^inftyspacefrac2^+nx^2^+n+1,=frac1x-1qquadqquadqquadqquadqquadtag2 $$
$$ S,,=sum_-infty^+inftyspacefrac2^n,x^2^n,,,+1,=S_small-+S_small+=frac1logxqquadqquadqquadtag3 $$
And find the convergence range of each sum.
In a similar question, the OP start by introducing $,(2),$ as a known identity, and gives a turly remarkable proof in the comment:
$$ sum_n=0^mlog(x^2^n+y^2^n) =logprod_n=0^m(x^2^n+y^2^n) =logfrac(x-y)x-yprod_n=0^m(x^2^n+y^2^n) =logfracx^2^m+1-y^2^m+1x-y $$
Then differentiate with respect to $y$, substitute $y=1$, and limit $,mtoinfty,$ give the result.
Could this approach work to proof $,(1),$? notice for $x=1$:
$$ S_small-=sum_n=1^inftyfrac12^n+1,=frac12 qquad=qquad lim_small xto1left[frac1logx-frac1x-1right]=frac12 $$
New alternative methods are highly welcomed. Thanks.
sequences-and-series summation power-series
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
edited Aug 27 at 3:09
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
asked Aug 26 at 19:27
Hazem Orabi
2,4812528
2,4812528
You might consider stopping altogether to play with the sizes of fonts, the result being exhausting to read...
– Did
Aug 26 at 21:31
@Did Apologizing.
– Hazem Orabi
Aug 26 at 21:42
1
The more general identity exists: $$sum_n=0^m frac2^n y^2^nx^2^n+y^2^n = fracyx-y-frac2^m+1y^2^m+1x^2^m+1-y^2^m+1 $$ For $y=1$ and $m rightarrow +infty$ we get what you desire.
– Tolaso
Aug 26 at 21:45
@Tolaso Again proofing (2). Thanks.
– Hazem Orabi
Aug 26 at 21:51
add a comment |Â
You might consider stopping altogether to play with the sizes of fonts, the result being exhausting to read...
– Did
Aug 26 at 21:31
@Did Apologizing.
– Hazem Orabi
Aug 26 at 21:42
1
The more general identity exists: $$sum_n=0^m frac2^n y^2^nx^2^n+y^2^n = fracyx-y-frac2^m+1y^2^m+1x^2^m+1-y^2^m+1 $$ For $y=1$ and $m rightarrow +infty$ we get what you desire.
– Tolaso
Aug 26 at 21:45
@Tolaso Again proofing (2). Thanks.
– Hazem Orabi
Aug 26 at 21:51
You might consider stopping altogether to play with the sizes of fonts, the result being exhausting to read...
– Did
Aug 26 at 21:31
You might consider stopping altogether to play with the sizes of fonts, the result being exhausting to read...
– Did
Aug 26 at 21:31
@Did Apologizing.
– Hazem Orabi
Aug 26 at 21:42
@Did Apologizing.
– Hazem Orabi
Aug 26 at 21:42
1
1
The more general identity exists: $$sum_n=0^m frac2^n y^2^nx^2^n+y^2^n = fracyx-y-frac2^m+1y^2^m+1x^2^m+1-y^2^m+1 $$ For $y=1$ and $m rightarrow +infty$ we get what you desire.
– Tolaso
Aug 26 at 21:45
The more general identity exists: $$sum_n=0^m frac2^n y^2^nx^2^n+y^2^n = fracyx-y-frac2^m+1y^2^m+1x^2^m+1-y^2^m+1 $$ For $y=1$ and $m rightarrow +infty$ we get what you desire.
– Tolaso
Aug 26 at 21:45
@Tolaso Again proofing (2). Thanks.
– Hazem Orabi
Aug 26 at 21:51
@Tolaso Again proofing (2). Thanks.
– Hazem Orabi
Aug 26 at 21:51
add a comment |Â
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You might consider stopping altogether to play with the sizes of fonts, the result being exhausting to read...
– Did
Aug 26 at 21:31
@Did Apologizing.
– Hazem Orabi
Aug 26 at 21:42
1
The more general identity exists: $$sum_n=0^m frac2^n y^2^nx^2^n+y^2^n = fracyx-y-frac2^m+1y^2^m+1x^2^m+1-y^2^m+1 $$ For $y=1$ and $m rightarrow +infty$ we get what you desire.
– Tolaso
Aug 26 at 21:45
@Tolaso Again proofing (2). Thanks.
– Hazem Orabi
Aug 26 at 21:51