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What's the limit of the series $log_2(1-x)+x+x^2+x^4+x^8+cdots$. [closed]
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13
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Find
$$lim_xto1^-log_2(1-x)+x+x^2+x^4+x^8+cdots$$
I have found $1-dfrac1ln2$ as a lower bound, but not further than that
calculus sequences-and-series limits
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
asked Jul 28 at 4:47
Kapil Pause
693
closed as off-topic by Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, user223391 Jul 31 at 1:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, Community
add a comment |Â
up vote
13
down vote
favorite
Find
$$lim_xto1^-log_2(1-x)+x+x^2+x^4+x^8+cdots$$
I have found $1-dfrac1ln2$ as a lower bound, but not further than that
calculus sequences-and-series limits
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
asked Jul 28 at 4:47
Kapil Pause
693
closed as off-topic by Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, user223391 Jul 31 at 1:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, Community
Does the limit exist?
– Szeto
Jul 28 at 5:51
Most probably yes....it's a decreasing function bounded from below.
– Kapil Pause
Jul 28 at 7:27
What does the last part, $$sum_k=0^infty x^2^k$$ alone give ($|x| < 1$)?
– Jeppe Stig Nielsen
Jul 28 at 8:38
@KapilPause: what is the bound ?
– Yves Daoust
Jul 28 at 15:10
How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn♦
Jul 31 at 18:03
add a comment |Â
up vote
13
down vote
favorite
up vote
13
down vote
favorite
Find
$$lim_xto1^-log_2(1-x)+x+x^2+x^4+x^8+cdots$$
I have found $1-dfrac1ln2$ as a lower bound, but not further than that
calculus sequences-and-series limits
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
asked Jul 28 at 4:47
Kapil Pause
693
Find
$$lim_xto1^-log_2(1-x)+x+x^2+x^4+x^8+cdots$$
I have found $1-dfrac1ln2$ as a lower bound, but not further than that
calculus sequences-and-series limits
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
asked Jul 28 at 4:47
Kapil Pause
693
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
edited Jul 30 at 19:30
Lorenzo B.
1,5402318
1,5402318
asked Jul 28 at 4:47
Kapil Pause
693
asked Jul 28 at 4:47
Kapil Pause
693
asked Jul 28 at 4:47
Kapil Pause
693
693
closed as off-topic by Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, user223391 Jul 31 at 1:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, Community
closed as off-topic by Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, user223391 Jul 31 at 1:08
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Carl Mummert, Xander Henderson, amWhy, Mostafa Ayaz, Community
Does the limit exist?
– Szeto
Jul 28 at 5:51
Most probably yes....it's a decreasing function bounded from below.
– Kapil Pause
Jul 28 at 7:27
What does the last part, $$sum_k=0^infty x^2^k$$ alone give ($|x| < 1$)?
– Jeppe Stig Nielsen
Jul 28 at 8:38
@KapilPause: what is the bound ?
– Yves Daoust
Jul 28 at 15:10
How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn♦
Jul 31 at 18:03
add a comment |Â
Does the limit exist?
– Szeto
Jul 28 at 5:51
Most probably yes....it's a decreasing function bounded from below.
– Kapil Pause
Jul 28 at 7:27
What does the last part, $$sum_k=0^infty x^2^k$$ alone give ($|x| < 1$)?
– Jeppe Stig Nielsen
Jul 28 at 8:38
@KapilPause: what is the bound ?
– Yves Daoust
Jul 28 at 15:10
How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn♦
Jul 31 at 18:03
Does the limit exist?
– Szeto
Jul 28 at 5:51
Does the limit exist?
– Szeto
Jul 28 at 5:51
Most probably yes....it's a decreasing function bounded from below.
– Kapil Pause
Jul 28 at 7:27
Most probably yes....it's a decreasing function bounded from below.
– Kapil Pause
Jul 28 at 7:27
What does the last part, $$sum_k=0^infty x^2^k$$ alone give ($|x| < 1$)?
– Jeppe Stig Nielsen
Jul 28 at 8:38
What does the last part, $$sum_k=0^infty x^2^k$$ alone give ($|x| < 1$)?
– Jeppe Stig Nielsen
Jul 28 at 8:38
@KapilPause: what is the bound ?
– Yves Daoust
Jul 28 at 15:10
@KapilPause: what is the bound ?
– Yves Daoust
Jul 28 at 15:10
How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn♦
Jul 31 at 18:03
How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn♦
Jul 31 at 18:03
add a comment |Â
3 Answers
3
active
oldest
votes
up vote
8
down vote
Let
$$
f(x)=log_2(1-x)+sum_k=0^infty x^2^ktag1
$$
then $f(0)=0$ and
$$
f!left(x^2right)=log_2left(1-x^2right)+sum_k=1^infty x^2^ktag2
$$
and therefore,
$$
f(x)-f!left(x^2right)=x-log_2(1+x)tag3
$$
Thus, for $xin(0,1)$,
$$
beginalign
f(1)
&=f(1)-f(0)\[12pt]
&=sum_k=-infty^inftyleft[f!left(x^2^kright)-f!left(x^2^k+1right)right]\
&=sum_k=-infty^inftyleft(x^2^k-log_2left(1+x^2^kright)right)tag4
endalign
$$
Expanding $log(1+x)$ into its Taylor Series in $x$, we get
$$
beginalign
int_0^1x^a-1log(1+x),mathrmdx
&=int_0^1sum_k=1^inftyfrac(-1)^k-1x^a-1+kk,mathrmdx\
&=sum_k=1^inftyfrac(-1)^k-1k(k+a)\
&=frac1asum_k=1^infty(-1)^k-1left(frac1k-frac1k+aright)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-2sum_k=1^inftyleft(frac12k-frac12k+aright)right)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-sum_k=1^inftyleft(frac1k-frac1k+a/2right)right)\[3pt]
&=fracH(a)-H(a/2)atag5
endalign
$$
where $H(a)$ are the Extended Harmonic Numbers. Apply $(5)$ to get
$$
int_0^1log_2left(1+x^2^kright),mathrmdx=fracH!left(2^-kright)-H!left(2^-k-1right)log(2)tag6
$$
Integration of a monomial gives
$$
int_0^1x^2^k,mathrmdx=frac12^k+1tag7
$$
Integrating $(4)$ over $[0,1]$ and using $(6)$ and $(7)$ yields
$$
beginalign
f(1)
&=lim_ntoinftyleft[sum_k=-n^nfrac12^k+1-fracH!left(2^nright)-H!left(2^-n-1right)log(2)right]\
&=lim_ntoinftyleft[frac12+n-fracgamma+nlog(2)+O!left(2^-nright)log(2)right]\[3pt]
&=frac12-fracgammalog(2)tag8
endalign
$$
Problem with the Use of Equation $boldsymbol(3)$
As pointed out by Michael, the use of equation $(3)$ above ignores the fact that
$$
g(x)-g!left(x^2right)=0tag9
$$
does not mean $g(x)=0$. In fact, for any $1$-periodic $h$, i.e. $h(x)=h(x+1)$,
$$
g(x)=h!left(log_2(-log(x))right)tag10
$$
satisfies $(9)$. I have encountered this misbehavior before in Does the family of series have a limit? and Find $f'(0)$ if $f(x)+f(2x)=xspacespaceforall x$.
Thus, the value given in $(8)$ is an average of the values of $f(1)$ given by $(4)$.
The function given in $(4)$ for $x=2^-2^-t$ has period $1$ in $t$. I have computed $f(1)$ from $(4)$ for $xinleft[frac14,frac12right]$; that is, the full period $tin[-1,0]$. I get a plot very similar to that of Michael:
which oscillates between $-0.33274775$ and $-0.33274460$. The horizontal line is
$$
frac12-fracgammalog(2)=-0.33274618
$$
which is pretty close to the average of the minimum and maximum.
I am still looking for an a priori method to compute this oscillation.
answered Jul 29 at 5:40
robjohn♦
258k25296612
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
 |Â
show 1 more comment
up vote
5
down vote
$$f(x)+log_2(1+x)-x=f(x^2)text ( 0<x<1 )\
f(exp(y))+log_2(1+exp(y))-exp(y)=f(exp(2y))text $(-infty<y<0)$\
g(y)+log_2(1+exp(y))-exp(y)=g(2y)text $(-infty<y<0)$\
g(-2^z)+log_2(1+exp(-2^z))-exp(-2^z)=g(-2^z+1)text( $-infty<z<infty$)\
h(z)+log_2(1+exp(-2^z))-exp(-2^z)=h(z+1)text$(-infty<z<infty)$\
f(1)-f(0)=int_-infty^inftydh=int_-infty^inftylog_2(1+exp(-2^z))-exp(-2^z)dz\
approx -0.332746$$
By change of variable, it becomes
$$int_0^1 fracx-log_2(1+x)xlog xlog 2dx$$
which the Inverse Symbolic Calculator gives as
$$frac12-fracgammaln2approx -0.3327461772769$$
As pointed out by Somos, I took an approximation when I replaced
$sum h(z+1)-h(z)$ by $int dh$. It seems to have variation in the sixth decimal place as $x$ varies from $x_0$ to $x_0^2$.
answered Jul 28 at 7:39
Michael
2,577213
add a comment |Â
up vote
2
down vote
Let $ f(x) := log_2(1-x) + sum_n=0^infty x^2^n, $
$ g(x) := f(e^-x) = log_2(1-e^-x) + sum_n=0^infty e^-x2^n, $
and $ a_k := g(2^-k) = b_k +
sum_n=0^infty e^-2^n-k $ where
$ b_k := log_2(1-e^-2^-k) approx -k - 2^-1-k/log(2). $
Now $ sum_n=0^infty e^-2^n-k = sum_n=1^k e^-2^-n + B $
where $ B := sum_n=0^infty e^-2^n approx 0.521865938459879089046726. $
But $ c_k :=! -k !+! sum_n=1^k e^-2^-n! =
sum_n=1^k big(e^-2^-n!-!1 big) $
and $ c_k to C $ where $ C approx -0.8546133208927. $
Finally, $ lim_xto 1^- f(x) = lim_xto 0^+ g(x) =
lim_ktoinfty a_k = B+C approx -0.3327473824328992250. $
The digits of $1$ minus this number is OEIS sequence A158468. $f(exp(-2^-30)) approx -0.3327473822.$
EDIT: Unfortunately, it seems that the function $ f(x) $ oscillates as it gets close to $1$ from below. That is, $ g(2^-x) $ approaches a period $1$ function with mean value $ 1/2 - gamma/log(2) $ with oscillations of magnitude $ approx 1.57315times 10^-6 $ as Michael shows. Thus, the limit does not exist. It was obvious that the infinite sum in $ f(x) $ has radius of convergence $1.$ What was not obvious was the limiting behavior as $ xto 1^-. $ We now know that the series has a logarithmic singularity and $ f(x) $ is what remains. That $ f(x) $ has interesting oscillatory behavior is nice information.
answered Jul 28 at 14:49
Somos
11k1831
add a comment |Â
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
8
down vote
Let
$$
f(x)=log_2(1-x)+sum_k=0^infty x^2^ktag1
$$
then $f(0)=0$ and
$$
f!left(x^2right)=log_2left(1-x^2right)+sum_k=1^infty x^2^ktag2
$$
and therefore,
$$
f(x)-f!left(x^2right)=x-log_2(1+x)tag3
$$
Thus, for $xin(0,1)$,
$$
beginalign
f(1)
&=f(1)-f(0)\[12pt]
&=sum_k=-infty^inftyleft[f!left(x^2^kright)-f!left(x^2^k+1right)right]\
&=sum_k=-infty^inftyleft(x^2^k-log_2left(1+x^2^kright)right)tag4
endalign
$$
Expanding $log(1+x)$ into its Taylor Series in $x$, we get
$$
beginalign
int_0^1x^a-1log(1+x),mathrmdx
&=int_0^1sum_k=1^inftyfrac(-1)^k-1x^a-1+kk,mathrmdx\
&=sum_k=1^inftyfrac(-1)^k-1k(k+a)\
&=frac1asum_k=1^infty(-1)^k-1left(frac1k-frac1k+aright)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-2sum_k=1^inftyleft(frac12k-frac12k+aright)right)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-sum_k=1^inftyleft(frac1k-frac1k+a/2right)right)\[3pt]
&=fracH(a)-H(a/2)atag5
endalign
$$
where $H(a)$ are the Extended Harmonic Numbers. Apply $(5)$ to get
$$
int_0^1log_2left(1+x^2^kright),mathrmdx=fracH!left(2^-kright)-H!left(2^-k-1right)log(2)tag6
$$
Integration of a monomial gives
$$
int_0^1x^2^k,mathrmdx=frac12^k+1tag7
$$
Integrating $(4)$ over $[0,1]$ and using $(6)$ and $(7)$ yields
$$
beginalign
f(1)
&=lim_ntoinftyleft[sum_k=-n^nfrac12^k+1-fracH!left(2^nright)-H!left(2^-n-1right)log(2)right]\
&=lim_ntoinftyleft[frac12+n-fracgamma+nlog(2)+O!left(2^-nright)log(2)right]\[3pt]
&=frac12-fracgammalog(2)tag8
endalign
$$
Problem with the Use of Equation $boldsymbol(3)$
As pointed out by Michael, the use of equation $(3)$ above ignores the fact that
$$
g(x)-g!left(x^2right)=0tag9
$$
does not mean $g(x)=0$. In fact, for any $1$-periodic $h$, i.e. $h(x)=h(x+1)$,
$$
g(x)=h!left(log_2(-log(x))right)tag10
$$
satisfies $(9)$. I have encountered this misbehavior before in Does the family of series have a limit? and Find $f'(0)$ if $f(x)+f(2x)=xspacespaceforall x$.
Thus, the value given in $(8)$ is an average of the values of $f(1)$ given by $(4)$.
The function given in $(4)$ for $x=2^-2^-t$ has period $1$ in $t$. I have computed $f(1)$ from $(4)$ for $xinleft[frac14,frac12right]$; that is, the full period $tin[-1,0]$. I get a plot very similar to that of Michael:
which oscillates between $-0.33274775$ and $-0.33274460$. The horizontal line is
$$
frac12-fracgammalog(2)=-0.33274618
$$
which is pretty close to the average of the minimum and maximum.
I am still looking for an a priori method to compute this oscillation.
answered Jul 29 at 5:40
robjohn♦
258k25296612
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
 |Â
show 1 more comment
up vote
8
down vote
Let
$$
f(x)=log_2(1-x)+sum_k=0^infty x^2^ktag1
$$
then $f(0)=0$ and
$$
f!left(x^2right)=log_2left(1-x^2right)+sum_k=1^infty x^2^ktag2
$$
and therefore,
$$
f(x)-f!left(x^2right)=x-log_2(1+x)tag3
$$
Thus, for $xin(0,1)$,
$$
beginalign
f(1)
&=f(1)-f(0)\[12pt]
&=sum_k=-infty^inftyleft[f!left(x^2^kright)-f!left(x^2^k+1right)right]\
&=sum_k=-infty^inftyleft(x^2^k-log_2left(1+x^2^kright)right)tag4
endalign
$$
Expanding $log(1+x)$ into its Taylor Series in $x$, we get
$$
beginalign
int_0^1x^a-1log(1+x),mathrmdx
&=int_0^1sum_k=1^inftyfrac(-1)^k-1x^a-1+kk,mathrmdx\
&=sum_k=1^inftyfrac(-1)^k-1k(k+a)\
&=frac1asum_k=1^infty(-1)^k-1left(frac1k-frac1k+aright)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-2sum_k=1^inftyleft(frac12k-frac12k+aright)right)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-sum_k=1^inftyleft(frac1k-frac1k+a/2right)right)\[3pt]
&=fracH(a)-H(a/2)atag5
endalign
$$
where $H(a)$ are the Extended Harmonic Numbers. Apply $(5)$ to get
$$
int_0^1log_2left(1+x^2^kright),mathrmdx=fracH!left(2^-kright)-H!left(2^-k-1right)log(2)tag6
$$
Integration of a monomial gives
$$
int_0^1x^2^k,mathrmdx=frac12^k+1tag7
$$
Integrating $(4)$ over $[0,1]$ and using $(6)$ and $(7)$ yields
$$
beginalign
f(1)
&=lim_ntoinftyleft[sum_k=-n^nfrac12^k+1-fracH!left(2^nright)-H!left(2^-n-1right)log(2)right]\
&=lim_ntoinftyleft[frac12+n-fracgamma+nlog(2)+O!left(2^-nright)log(2)right]\[3pt]
&=frac12-fracgammalog(2)tag8
endalign
$$
Problem with the Use of Equation $boldsymbol(3)$
As pointed out by Michael, the use of equation $(3)$ above ignores the fact that
$$
g(x)-g!left(x^2right)=0tag9
$$
does not mean $g(x)=0$. In fact, for any $1$-periodic $h$, i.e. $h(x)=h(x+1)$,
$$
g(x)=h!left(log_2(-log(x))right)tag10
$$
satisfies $(9)$. I have encountered this misbehavior before in Does the family of series have a limit? and Find $f'(0)$ if $f(x)+f(2x)=xspacespaceforall x$.
Thus, the value given in $(8)$ is an average of the values of $f(1)$ given by $(4)$.
The function given in $(4)$ for $x=2^-2^-t$ has period $1$ in $t$. I have computed $f(1)$ from $(4)$ for $xinleft[frac14,frac12right]$; that is, the full period $tin[-1,0]$. I get a plot very similar to that of Michael:
which oscillates between $-0.33274775$ and $-0.33274460$. The horizontal line is
$$
frac12-fracgammalog(2)=-0.33274618
$$
which is pretty close to the average of the minimum and maximum.
I am still looking for an a priori method to compute this oscillation.
answered Jul 29 at 5:40
robjohn♦
258k25296612
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
 |Â
show 1 more comment
up vote
8
down vote
up vote
8
down vote
Let
$$
f(x)=log_2(1-x)+sum_k=0^infty x^2^ktag1
$$
then $f(0)=0$ and
$$
f!left(x^2right)=log_2left(1-x^2right)+sum_k=1^infty x^2^ktag2
$$
and therefore,
$$
f(x)-f!left(x^2right)=x-log_2(1+x)tag3
$$
Thus, for $xin(0,1)$,
$$
beginalign
f(1)
&=f(1)-f(0)\[12pt]
&=sum_k=-infty^inftyleft[f!left(x^2^kright)-f!left(x^2^k+1right)right]\
&=sum_k=-infty^inftyleft(x^2^k-log_2left(1+x^2^kright)right)tag4
endalign
$$
Expanding $log(1+x)$ into its Taylor Series in $x$, we get
$$
beginalign
int_0^1x^a-1log(1+x),mathrmdx
&=int_0^1sum_k=1^inftyfrac(-1)^k-1x^a-1+kk,mathrmdx\
&=sum_k=1^inftyfrac(-1)^k-1k(k+a)\
&=frac1asum_k=1^infty(-1)^k-1left(frac1k-frac1k+aright)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-2sum_k=1^inftyleft(frac12k-frac12k+aright)right)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-sum_k=1^inftyleft(frac1k-frac1k+a/2right)right)\[3pt]
&=fracH(a)-H(a/2)atag5
endalign
$$
where $H(a)$ are the Extended Harmonic Numbers. Apply $(5)$ to get
$$
int_0^1log_2left(1+x^2^kright),mathrmdx=fracH!left(2^-kright)-H!left(2^-k-1right)log(2)tag6
$$
Integration of a monomial gives
$$
int_0^1x^2^k,mathrmdx=frac12^k+1tag7
$$
Integrating $(4)$ over $[0,1]$ and using $(6)$ and $(7)$ yields
$$
beginalign
f(1)
&=lim_ntoinftyleft[sum_k=-n^nfrac12^k+1-fracH!left(2^nright)-H!left(2^-n-1right)log(2)right]\
&=lim_ntoinftyleft[frac12+n-fracgamma+nlog(2)+O!left(2^-nright)log(2)right]\[3pt]
&=frac12-fracgammalog(2)tag8
endalign
$$
Problem with the Use of Equation $boldsymbol(3)$
As pointed out by Michael, the use of equation $(3)$ above ignores the fact that
$$
g(x)-g!left(x^2right)=0tag9
$$
does not mean $g(x)=0$. In fact, for any $1$-periodic $h$, i.e. $h(x)=h(x+1)$,
$$
g(x)=h!left(log_2(-log(x))right)tag10
$$
satisfies $(9)$. I have encountered this misbehavior before in Does the family of series have a limit? and Find $f'(0)$ if $f(x)+f(2x)=xspacespaceforall x$.
Thus, the value given in $(8)$ is an average of the values of $f(1)$ given by $(4)$.
The function given in $(4)$ for $x=2^-2^-t$ has period $1$ in $t$. I have computed $f(1)$ from $(4)$ for $xinleft[frac14,frac12right]$; that is, the full period $tin[-1,0]$. I get a plot very similar to that of Michael:
which oscillates between $-0.33274775$ and $-0.33274460$. The horizontal line is
$$
frac12-fracgammalog(2)=-0.33274618
$$
which is pretty close to the average of the minimum and maximum.
I am still looking for an a priori method to compute this oscillation.
answered Jul 29 at 5:40
robjohn♦
258k25296612
Let
$$
f(x)=log_2(1-x)+sum_k=0^infty x^2^ktag1
$$
then $f(0)=0$ and
$$
f!left(x^2right)=log_2left(1-x^2right)+sum_k=1^infty x^2^ktag2
$$
and therefore,
$$
f(x)-f!left(x^2right)=x-log_2(1+x)tag3
$$
Thus, for $xin(0,1)$,
$$
beginalign
f(1)
&=f(1)-f(0)\[12pt]
&=sum_k=-infty^inftyleft[f!left(x^2^kright)-f!left(x^2^k+1right)right]\
&=sum_k=-infty^inftyleft(x^2^k-log_2left(1+x^2^kright)right)tag4
endalign
$$
Expanding $log(1+x)$ into its Taylor Series in $x$, we get
$$
beginalign
int_0^1x^a-1log(1+x),mathrmdx
&=int_0^1sum_k=1^inftyfrac(-1)^k-1x^a-1+kk,mathrmdx\
&=sum_k=1^inftyfrac(-1)^k-1k(k+a)\
&=frac1asum_k=1^infty(-1)^k-1left(frac1k-frac1k+aright)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-2sum_k=1^inftyleft(frac12k-frac12k+aright)right)\
&=frac1aleft(sum_k=1^inftyleft(frac1k-frac1k+aright)-sum_k=1^inftyleft(frac1k-frac1k+a/2right)right)\[3pt]
&=fracH(a)-H(a/2)atag5
endalign
$$
where $H(a)$ are the Extended Harmonic Numbers. Apply $(5)$ to get
$$
int_0^1log_2left(1+x^2^kright),mathrmdx=fracH!left(2^-kright)-H!left(2^-k-1right)log(2)tag6
$$
Integration of a monomial gives
$$
int_0^1x^2^k,mathrmdx=frac12^k+1tag7
$$
Integrating $(4)$ over $[0,1]$ and using $(6)$ and $(7)$ yields
$$
beginalign
f(1)
&=lim_ntoinftyleft[sum_k=-n^nfrac12^k+1-fracH!left(2^nright)-H!left(2^-n-1right)log(2)right]\
&=lim_ntoinftyleft[frac12+n-fracgamma+nlog(2)+O!left(2^-nright)log(2)right]\[3pt]
&=frac12-fracgammalog(2)tag8
endalign
$$
Problem with the Use of Equation $boldsymbol(3)$
As pointed out by Michael, the use of equation $(3)$ above ignores the fact that
$$
g(x)-g!left(x^2right)=0tag9
$$
does not mean $g(x)=0$. In fact, for any $1$-periodic $h$, i.e. $h(x)=h(x+1)$,
$$
g(x)=h!left(log_2(-log(x))right)tag10
$$
satisfies $(9)$. I have encountered this misbehavior before in Does the family of series have a limit? and Find $f'(0)$ if $f(x)+f(2x)=xspacespaceforall x$.
Thus, the value given in $(8)$ is an average of the values of $f(1)$ given by $(4)$.
The function given in $(4)$ for $x=2^-2^-t$ has period $1$ in $t$. I have computed $f(1)$ from $(4)$ for $xinleft[frac14,frac12right]$; that is, the full period $tin[-1,0]$. I get a plot very similar to that of Michael:
which oscillates between $-0.33274775$ and $-0.33274460$. The horizontal line is
$$
frac12-fracgammalog(2)=-0.33274618
$$
which is pretty close to the average of the minimum and maximum.
I am still looking for an a priori method to compute this oscillation.
answered Jul 29 at 5:40
robjohn♦
258k25296612
edited Aug 4 at 21:05
answered Jul 29 at 5:40
robjohn♦
258k25296612
answered Jul 29 at 5:40
robjohn♦
258k25296612
answered Jul 29 at 5:40
robjohn♦
258k25296612
258k25296612
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
 |Â
show 1 more comment
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
When you integrate (4), you average over the possible values of $f(1)$. My graph says $f(x)$ oscillates near $x=1$ with amplitude 0.0000015. Is that true, or a numerical artifact of underflow ?
– Michael
Jul 29 at 7:32
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
@Michael: if we differentiate $(3)$, and evaluate as $xto1^-$, we get $f'(1)=frac12log(2)-1$; so the function apparently decreases to $frac12-fracgammalog(2)$ as $xto1^-$.
– robjohn♦
Jul 29 at 12:09
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
The periodicity is that $f(x)=p(x)+q(x)$ where $q(x)=q(x^2)$, for example $q(x)=sin(2pilog_2(-log(x)))$. So $q(x)$ is cancelled in (3), which therefore can't rule out $q(x)$
– Michael
Jul 29 at 12:24
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
@Michael: ah, yes. I have encountered this before, and I was worried about it because of $(4)$. Equation $(3)$ kills the effect of any such $q$. I will have to think on this and try to see if I can compute this component.
– robjohn♦
Jul 29 at 12:36
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
It's not exact, but very close with $x=2^-2^-t$: $$ frac12 - fracgammaln(2) - fracleft(pi^pi + gamma^piright)^1/pi2cdot 10^-6 , sin(2pi t - 0.692) $$
– Diger
Jul 30 at 22:59
 |Â
show 1 more comment
up vote
5
down vote
$$f(x)+log_2(1+x)-x=f(x^2)text ( 0<x<1 )\
f(exp(y))+log_2(1+exp(y))-exp(y)=f(exp(2y))text $(-infty<y<0)$\
g(y)+log_2(1+exp(y))-exp(y)=g(2y)text $(-infty<y<0)$\
g(-2^z)+log_2(1+exp(-2^z))-exp(-2^z)=g(-2^z+1)text( $-infty<z<infty$)\
h(z)+log_2(1+exp(-2^z))-exp(-2^z)=h(z+1)text$(-infty<z<infty)$\
f(1)-f(0)=int_-infty^inftydh=int_-infty^inftylog_2(1+exp(-2^z))-exp(-2^z)dz\
approx -0.332746$$
By change of variable, it becomes
$$int_0^1 fracx-log_2(1+x)xlog xlog 2dx$$
which the Inverse Symbolic Calculator gives as
$$frac12-fracgammaln2approx -0.3327461772769$$
As pointed out by Somos, I took an approximation when I replaced
$sum h(z+1)-h(z)$ by $int dh$. It seems to have variation in the sixth decimal place as $x$ varies from $x_0$ to $x_0^2$.
answered Jul 28 at 7:39
Michael
2,577213
add a comment |Â
up vote
5
down vote
$$f(x)+log_2(1+x)-x=f(x^2)text ( 0<x<1 )\
f(exp(y))+log_2(1+exp(y))-exp(y)=f(exp(2y))text $(-infty<y<0)$\
g(y)+log_2(1+exp(y))-exp(y)=g(2y)text $(-infty<y<0)$\
g(-2^z)+log_2(1+exp(-2^z))-exp(-2^z)=g(-2^z+1)text( $-infty<z<infty$)\
h(z)+log_2(1+exp(-2^z))-exp(-2^z)=h(z+1)text$(-infty<z<infty)$\
f(1)-f(0)=int_-infty^inftydh=int_-infty^inftylog_2(1+exp(-2^z))-exp(-2^z)dz\
approx -0.332746$$
By change of variable, it becomes
$$int_0^1 fracx-log_2(1+x)xlog xlog 2dx$$
which the Inverse Symbolic Calculator gives as
$$frac12-fracgammaln2approx -0.3327461772769$$
As pointed out by Somos, I took an approximation when I replaced
$sum h(z+1)-h(z)$ by $int dh$. It seems to have variation in the sixth decimal place as $x$ varies from $x_0$ to $x_0^2$.
answered Jul 28 at 7:39
Michael
2,577213
add a comment |Â
up vote
5
down vote
up vote
5
down vote
$$f(x)+log_2(1+x)-x=f(x^2)text ( 0<x<1 )\
f(exp(y))+log_2(1+exp(y))-exp(y)=f(exp(2y))text $(-infty<y<0)$\
g(y)+log_2(1+exp(y))-exp(y)=g(2y)text $(-infty<y<0)$\
g(-2^z)+log_2(1+exp(-2^z))-exp(-2^z)=g(-2^z+1)text( $-infty<z<infty$)\
h(z)+log_2(1+exp(-2^z))-exp(-2^z)=h(z+1)text$(-infty<z<infty)$\
f(1)-f(0)=int_-infty^inftydh=int_-infty^inftylog_2(1+exp(-2^z))-exp(-2^z)dz\
approx -0.332746$$
By change of variable, it becomes
$$int_0^1 fracx-log_2(1+x)xlog xlog 2dx$$
which the Inverse Symbolic Calculator gives as
$$frac12-fracgammaln2approx -0.3327461772769$$
As pointed out by Somos, I took an approximation when I replaced
$sum h(z+1)-h(z)$ by $int dh$. It seems to have variation in the sixth decimal place as $x$ varies from $x_0$ to $x_0^2$.
answered Jul 28 at 7:39
Michael
2,577213
$$f(x)+log_2(1+x)-x=f(x^2)text ( 0<x<1 )\
f(exp(y))+log_2(1+exp(y))-exp(y)=f(exp(2y))text $(-infty<y<0)$\
g(y)+log_2(1+exp(y))-exp(y)=g(2y)text $(-infty<y<0)$\
g(-2^z)+log_2(1+exp(-2^z))-exp(-2^z)=g(-2^z+1)text( $-infty<z<infty$)\
h(z)+log_2(1+exp(-2^z))-exp(-2^z)=h(z+1)text$(-infty<z<infty)$\
f(1)-f(0)=int_-infty^inftydh=int_-infty^inftylog_2(1+exp(-2^z))-exp(-2^z)dz\
approx -0.332746$$
By change of variable, it becomes
$$int_0^1 fracx-log_2(1+x)xlog xlog 2dx$$
which the Inverse Symbolic Calculator gives as
$$frac12-fracgammaln2approx -0.3327461772769$$
As pointed out by Somos, I took an approximation when I replaced
$sum h(z+1)-h(z)$ by $int dh$. It seems to have variation in the sixth decimal place as $x$ varies from $x_0$ to $x_0^2$.
answered Jul 28 at 7:39
Michael
2,577213
edited Jul 29 at 2:08
answered Jul 28 at 7:39
Michael
2,577213
answered Jul 28 at 7:39
Michael
2,577213
answered Jul 28 at 7:39
Michael
2,577213
2,577213
add a comment |Â
add a comment |Â
up vote
2
down vote
Let $ f(x) := log_2(1-x) + sum_n=0^infty x^2^n, $
$ g(x) := f(e^-x) = log_2(1-e^-x) + sum_n=0^infty e^-x2^n, $
and $ a_k := g(2^-k) = b_k +
sum_n=0^infty e^-2^n-k $ where
$ b_k := log_2(1-e^-2^-k) approx -k - 2^-1-k/log(2). $
Now $ sum_n=0^infty e^-2^n-k = sum_n=1^k e^-2^-n + B $
where $ B := sum_n=0^infty e^-2^n approx 0.521865938459879089046726. $
But $ c_k :=! -k !+! sum_n=1^k e^-2^-n! =
sum_n=1^k big(e^-2^-n!-!1 big) $
and $ c_k to C $ where $ C approx -0.8546133208927. $
Finally, $ lim_xto 1^- f(x) = lim_xto 0^+ g(x) =
lim_ktoinfty a_k = B+C approx -0.3327473824328992250. $
The digits of $1$ minus this number is OEIS sequence A158468. $f(exp(-2^-30)) approx -0.3327473822.$
EDIT: Unfortunately, it seems that the function $ f(x) $ oscillates as it gets close to $1$ from below. That is, $ g(2^-x) $ approaches a period $1$ function with mean value $ 1/2 - gamma/log(2) $ with oscillations of magnitude $ approx 1.57315times 10^-6 $ as Michael shows. Thus, the limit does not exist. It was obvious that the infinite sum in $ f(x) $ has radius of convergence $1.$ What was not obvious was the limiting behavior as $ xto 1^-. $ We now know that the series has a logarithmic singularity and $ f(x) $ is what remains. That $ f(x) $ has interesting oscillatory behavior is nice information.
answered Jul 28 at 14:49
Somos
11k1831
add a comment |Â
up vote
2
down vote
Let $ f(x) := log_2(1-x) + sum_n=0^infty x^2^n, $
$ g(x) := f(e^-x) = log_2(1-e^-x) + sum_n=0^infty e^-x2^n, $
and $ a_k := g(2^-k) = b_k +
sum_n=0^infty e^-2^n-k $ where
$ b_k := log_2(1-e^-2^-k) approx -k - 2^-1-k/log(2). $
Now $ sum_n=0^infty e^-2^n-k = sum_n=1^k e^-2^-n + B $
where $ B := sum_n=0^infty e^-2^n approx 0.521865938459879089046726. $
But $ c_k :=! -k !+! sum_n=1^k e^-2^-n! =
sum_n=1^k big(e^-2^-n!-!1 big) $
and $ c_k to C $ where $ C approx -0.8546133208927. $
Finally, $ lim_xto 1^- f(x) = lim_xto 0^+ g(x) =
lim_ktoinfty a_k = B+C approx -0.3327473824328992250. $
The digits of $1$ minus this number is OEIS sequence A158468. $f(exp(-2^-30)) approx -0.3327473822.$
EDIT: Unfortunately, it seems that the function $ f(x) $ oscillates as it gets close to $1$ from below. That is, $ g(2^-x) $ approaches a period $1$ function with mean value $ 1/2 - gamma/log(2) $ with oscillations of magnitude $ approx 1.57315times 10^-6 $ as Michael shows. Thus, the limit does not exist. It was obvious that the infinite sum in $ f(x) $ has radius of convergence $1.$ What was not obvious was the limiting behavior as $ xto 1^-. $ We now know that the series has a logarithmic singularity and $ f(x) $ is what remains. That $ f(x) $ has interesting oscillatory behavior is nice information.
answered Jul 28 at 14:49
Somos
11k1831
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Let $ f(x) := log_2(1-x) + sum_n=0^infty x^2^n, $
$ g(x) := f(e^-x) = log_2(1-e^-x) + sum_n=0^infty e^-x2^n, $
and $ a_k := g(2^-k) = b_k +
sum_n=0^infty e^-2^n-k $ where
$ b_k := log_2(1-e^-2^-k) approx -k - 2^-1-k/log(2). $
Now $ sum_n=0^infty e^-2^n-k = sum_n=1^k e^-2^-n + B $
where $ B := sum_n=0^infty e^-2^n approx 0.521865938459879089046726. $
But $ c_k :=! -k !+! sum_n=1^k e^-2^-n! =
sum_n=1^k big(e^-2^-n!-!1 big) $
and $ c_k to C $ where $ C approx -0.8546133208927. $
Finally, $ lim_xto 1^- f(x) = lim_xto 0^+ g(x) =
lim_ktoinfty a_k = B+C approx -0.3327473824328992250. $
The digits of $1$ minus this number is OEIS sequence A158468. $f(exp(-2^-30)) approx -0.3327473822.$
EDIT: Unfortunately, it seems that the function $ f(x) $ oscillates as it gets close to $1$ from below. That is, $ g(2^-x) $ approaches a period $1$ function with mean value $ 1/2 - gamma/log(2) $ with oscillations of magnitude $ approx 1.57315times 10^-6 $ as Michael shows. Thus, the limit does not exist. It was obvious that the infinite sum in $ f(x) $ has radius of convergence $1.$ What was not obvious was the limiting behavior as $ xto 1^-. $ We now know that the series has a logarithmic singularity and $ f(x) $ is what remains. That $ f(x) $ has interesting oscillatory behavior is nice information.
answered Jul 28 at 14:49
Somos
11k1831
Let $ f(x) := log_2(1-x) + sum_n=0^infty x^2^n, $
$ g(x) := f(e^-x) = log_2(1-e^-x) + sum_n=0^infty e^-x2^n, $
and $ a_k := g(2^-k) = b_k +
sum_n=0^infty e^-2^n-k $ where
$ b_k := log_2(1-e^-2^-k) approx -k - 2^-1-k/log(2). $
Now $ sum_n=0^infty e^-2^n-k = sum_n=1^k e^-2^-n + B $
where $ B := sum_n=0^infty e^-2^n approx 0.521865938459879089046726. $
But $ c_k :=! -k !+! sum_n=1^k e^-2^-n! =
sum_n=1^k big(e^-2^-n!-!1 big) $
and $ c_k to C $ where $ C approx -0.8546133208927. $
Finally, $ lim_xto 1^- f(x) = lim_xto 0^+ g(x) =
lim_ktoinfty a_k = B+C approx -0.3327473824328992250. $
The digits of $1$ minus this number is OEIS sequence A158468. $f(exp(-2^-30)) approx -0.3327473822.$
EDIT: Unfortunately, it seems that the function $ f(x) $ oscillates as it gets close to $1$ from below. That is, $ g(2^-x) $ approaches a period $1$ function with mean value $ 1/2 - gamma/log(2) $ with oscillations of magnitude $ approx 1.57315times 10^-6 $ as Michael shows. Thus, the limit does not exist. It was obvious that the infinite sum in $ f(x) $ has radius of convergence $1.$ What was not obvious was the limiting behavior as $ xto 1^-. $ We now know that the series has a logarithmic singularity and $ f(x) $ is what remains. That $ f(x) $ has interesting oscillatory behavior is nice information.
answered Jul 28 at 14:49
Somos
11k1831
edited Jul 29 at 14:36
answered Jul 28 at 14:49
Somos
11k1831
answered Jul 28 at 14:49
Somos
11k1831
answered Jul 28 at 14:49
Somos
11k1831
11k1831
add a comment |Â
add a comment |Â
Does the limit exist?
– Szeto
Jul 28 at 5:51
Most probably yes....it's a decreasing function bounded from below.
– Kapil Pause
Jul 28 at 7:27
What does the last part, $$sum_k=0^infty x^2^k$$ alone give ($|x| < 1$)?
– Jeppe Stig Nielsen
Jul 28 at 8:38
@KapilPause: what is the bound ?
– Yves Daoust
Jul 28 at 15:10
How did you get that lower bound? Perhaps if you added that, the question might be reopened.
– robjohn♦
Jul 31 at 18:03