Mixing
Evaluation of $sumlimits_{k=1}^nleft(x^{k}+frac{1}{x^k}right)^k$
$begingroup$
Can anyone find a simplified expression for the sum $displaystyle sum_{k=1}^nleft(x^{k}+frac{1}{x^k}right)^k$? I have tried expanding the first few terms but it gets a little messy with no clear leads. I suspect formulae for geometric series may come into it somehow, but at the moment it isn't clear how to start.
sequences-and-series algebra-precalculus binomial-theorem
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
$endgroup$
|
show 2 more comments
$begingroup$
Can anyone find a simplified expression for the sum $displaystyle sum_{k=1}^nleft(x^{k}+frac{1}{x^k}right)^k$? I have tried expanding the first few terms but it gets a little messy with no clear leads. I suspect formulae for geometric series may come into it somehow, but at the moment it isn't clear how to start.
sequences-and-series algebra-precalculus binomial-theorem
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
$endgroup$
$begingroup$
It's from an Olympiad paper I believe, so I assumed there would be a neat solution, with no need for double sums etc.
$endgroup$
– wrb98
Mar 8 '17 at 20:46
2
$begingroup$
I honestly have no clue. After binomial expansion, you end up with things like $x^{k^2}$, which have no closed form sum.
$endgroup$
– Simply Beautiful Art
Mar 8 '17 at 20:47
$begingroup$
Exactly. I tried to see if expanding the first brackets could lead to some pattern and then proceed by induction, but it's a dead end.
$endgroup$
– wrb98
Mar 8 '17 at 20:49
1
$begingroup$
Interesting, even Wolfram|Alpha can't figure it out.
$endgroup$
– projectilemotion
Mar 8 '17 at 20:52
3
$begingroup$
@Will: is the original problem really about simplifying such sum, or just about evaluating the coefficient of $x^0$ in the Laurent series?
$endgroup$
– Jack D'Aurizio
Mar 8 '17 at 21:06
|
show 2 more comments
$begingroup$
Can anyone find a simplified expression for the sum $displaystyle sum_{k=1}^nleft(x^{k}+frac{1}{x^k}right)^k$? I have tried expanding the first few terms but it gets a little messy with no clear leads. I suspect formulae for geometric series may come into it somehow, but at the moment it isn't clear how to start.
sequences-and-series algebra-precalculus binomial-theorem
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
$endgroup$
Can anyone find a simplified expression for the sum $displaystyle sum_{k=1}^nleft(x^{k}+frac{1}{x^k}right)^k$? I have tried expanding the first few terms but it gets a little messy with no clear leads. I suspect formulae for geometric series may come into it somehow, but at the moment it isn't clear how to start.
sequences-and-series algebra-precalculus binomial-theorem
sequences-and-series algebra-precalculus binomial-theorem
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
edited Jan 18 at 12:51
Martin Sleziak
44.7k10119272
44.7k10119272
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
asked Mar 8 '17 at 20:44
wrb98wrb98
579313
579313
$begingroup$
It's from an Olympiad paper I believe, so I assumed there would be a neat solution, with no need for double sums etc.
$endgroup$
– wrb98
Mar 8 '17 at 20:46
2
$begingroup$
I honestly have no clue. After binomial expansion, you end up with things like $x^{k^2}$, which have no closed form sum.
$endgroup$
– Simply Beautiful Art
Mar 8 '17 at 20:47
$begingroup$
Exactly. I tried to see if expanding the first brackets could lead to some pattern and then proceed by induction, but it's a dead end.
$endgroup$
– wrb98
Mar 8 '17 at 20:49
1
$begingroup$
Interesting, even Wolfram|Alpha can't figure it out.
$endgroup$
– projectilemotion
Mar 8 '17 at 20:52
3
$begingroup$
@Will: is the original problem really about simplifying such sum, or just about evaluating the coefficient of $x^0$ in the Laurent series?
$endgroup$
– Jack D'Aurizio
Mar 8 '17 at 21:06
|
show 2 more comments
$begingroup$
It's from an Olympiad paper I believe, so I assumed there would be a neat solution, with no need for double sums etc.
$endgroup$
– wrb98
Mar 8 '17 at 20:46
2
$begingroup$
I honestly have no clue. After binomial expansion, you end up with things like $x^{k^2}$, which have no closed form sum.
$endgroup$
– Simply Beautiful Art
Mar 8 '17 at 20:47
$begingroup$
Exactly. I tried to see if expanding the first brackets could lead to some pattern and then proceed by induction, but it's a dead end.
$endgroup$
– wrb98
Mar 8 '17 at 20:49
1
$begingroup$
Interesting, even Wolfram|Alpha can't figure it out.
$endgroup$
– projectilemotion
Mar 8 '17 at 20:52
3
$begingroup$
@Will: is the original problem really about simplifying such sum, or just about evaluating the coefficient of $x^0$ in the Laurent series?
$endgroup$
– Jack D'Aurizio
Mar 8 '17 at 21:06
$begingroup$
It's from an Olympiad paper I believe, so I assumed there would be a neat solution, with no need for double sums etc.
$endgroup$
– wrb98
Mar 8 '17 at 20:46
$begingroup$
It's from an Olympiad paper I believe, so I assumed there would be a neat solution, with no need for double sums etc.
$endgroup$
– wrb98
Mar 8 '17 at 20:46
2
2
$begingroup$
I honestly have no clue. After binomial expansion, you end up with things like $x^{k^2}$, which have no closed form sum.
$endgroup$
– Simply Beautiful Art
Mar 8 '17 at 20:47
$begingroup$
I honestly have no clue. After binomial expansion, you end up with things like $x^{k^2}$, which have no closed form sum.
$endgroup$
– Simply Beautiful Art
Mar 8 '17 at 20:47
$begingroup$
Exactly. I tried to see if expanding the first brackets could lead to some pattern and then proceed by induction, but it's a dead end.
$endgroup$
– wrb98
Mar 8 '17 at 20:49
$begingroup$
Exactly. I tried to see if expanding the first brackets could lead to some pattern and then proceed by induction, but it's a dead end.
$endgroup$
– wrb98
Mar 8 '17 at 20:49
1
1
$begingroup$
Interesting, even Wolfram|Alpha can't figure it out.
$endgroup$
– projectilemotion
Mar 8 '17 at 20:52
$begingroup$
Interesting, even Wolfram|Alpha can't figure it out.
$endgroup$
– projectilemotion
Mar 8 '17 at 20:52
3
3
$begingroup$
@Will: is the original problem really about simplifying such sum, or just about evaluating the coefficient of $x^0$ in the Laurent series?
$endgroup$
– Jack D'Aurizio
Mar 8 '17 at 21:06
$begingroup$
@Will: is the original problem really about simplifying such sum, or just about evaluating the coefficient of $x^0$ in the Laurent series?
$endgroup$
– Jack D'Aurizio
Mar 8 '17 at 21:06
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
Let $S(n)$ be your sum.
For $1 le m le n^2$, the coefficient of $x^m$ (and of $x^{-m}$, by symmetry) in $S(n)$ is
$$ [x^m]; S(n) = sum_k {k choose frac{m}{2k}+frac{k}{2}}$$
where the sum is over all divisors $k$ of $m$ such that $m le k^2 le n^2$
and $frac{m}{k} equiv k mod 2$. In particular this is $0$ if $m equiv 2 mod 4$.
The coefficient of $x^0$ in $S(n)$ is $$ [x^0] ; S(n) = sum_{j=1}^{lfloor n/2 rfloor} {2j choose j}$$
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
1
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
add a comment |
$begingroup$
Not an answer, just a long comment;
Not sure how to proceed, but one way to rewrite it is $x=e^{iphi}$, getting
$$sum_{k=1}^n (2cos kphi)^k=sum_{k=1}^n 2^k T_k^k(cos(phi))$$
where $T$ is a Chebyshev polynomial of the first kind.
It seems to me that the expression is too convoluted to expect a nice solution. Another similar approach that might be considered is to express everything with $x+x^{-1}$. We have expressions of the type
$$(x+x^{-1})^2=2+(x^2+x^{-2})$$
$$(x+x^{-1})^3=3(x+x^{-1})+(x^3+x^{-3})$$
which could be solved for $x^k+x^{-k}$ but I can't see how to get a closed form for a general term. This solution does give you the constant term (reproduces result by @RobertIsrael) but the rest gets complicated due to $()^k$ - we get summations over very strange sets.
A reference that might be useful - it includes some very similar expressions.
answered Jan 18 at 14:26
orionorion
13.6k11837
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Let $S(n)$ be your sum.
For $1 le m le n^2$, the coefficient of $x^m$ (and of $x^{-m}$, by symmetry) in $S(n)$ is
$$ [x^m]; S(n) = sum_k {k choose frac{m}{2k}+frac{k}{2}}$$
where the sum is over all divisors $k$ of $m$ such that $m le k^2 le n^2$
and $frac{m}{k} equiv k mod 2$. In particular this is $0$ if $m equiv 2 mod 4$.
The coefficient of $x^0$ in $S(n)$ is $$ [x^0] ; S(n) = sum_{j=1}^{lfloor n/2 rfloor} {2j choose j}$$
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
1
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
add a comment |
$begingroup$
Let $S(n)$ be your sum.
For $1 le m le n^2$, the coefficient of $x^m$ (and of $x^{-m}$, by symmetry) in $S(n)$ is
$$ [x^m]; S(n) = sum_k {k choose frac{m}{2k}+frac{k}{2}}$$
where the sum is over all divisors $k$ of $m$ such that $m le k^2 le n^2$
and $frac{m}{k} equiv k mod 2$. In particular this is $0$ if $m equiv 2 mod 4$.
The coefficient of $x^0$ in $S(n)$ is $$ [x^0] ; S(n) = sum_{j=1}^{lfloor n/2 rfloor} {2j choose j}$$
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
$endgroup$
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
1
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
add a comment |
$begingroup$
Let $S(n)$ be your sum.
For $1 le m le n^2$, the coefficient of $x^m$ (and of $x^{-m}$, by symmetry) in $S(n)$ is
$$ [x^m]; S(n) = sum_k {k choose frac{m}{2k}+frac{k}{2}}$$
where the sum is over all divisors $k$ of $m$ such that $m le k^2 le n^2$
and $frac{m}{k} equiv k mod 2$. In particular this is $0$ if $m equiv 2 mod 4$.
The coefficient of $x^0$ in $S(n)$ is $$ [x^0] ; S(n) = sum_{j=1}^{lfloor n/2 rfloor} {2j choose j}$$
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
$endgroup$
Let $S(n)$ be your sum.
For $1 le m le n^2$, the coefficient of $x^m$ (and of $x^{-m}$, by symmetry) in $S(n)$ is
$$ [x^m]; S(n) = sum_k {k choose frac{m}{2k}+frac{k}{2}}$$
where the sum is over all divisors $k$ of $m$ such that $m le k^2 le n^2$
and $frac{m}{k} equiv k mod 2$. In particular this is $0$ if $m equiv 2 mod 4$.
The coefficient of $x^0$ in $S(n)$ is $$ [x^0] ; S(n) = sum_{j=1}^{lfloor n/2 rfloor} {2j choose j}$$
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
edited Mar 8 '17 at 21:37
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
answered Mar 8 '17 at 21:30
Robert IsraelRobert Israel
325k23214468
325k23214468
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
1
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
add a comment |
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
1
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
$begingroup$
My guess, based on the usual approximation to $binom{2j}{j}$, is that $sum_{j=1}^m binom{2j}{j} approx dfrac{4^m}{sqrt{pi }}dfrac{4}{3sqrt{m}}$.
$endgroup$
– marty cohen
Mar 9 '17 at 3:23
1
1
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
$begingroup$
@martycohen See OEIS sequence A066796, in particular the formula by Vaclav Kotesovec, which agrees with your guess.
$endgroup$
– Robert Israel
Mar 9 '17 at 4:53
add a comment |
$begingroup$
Not an answer, just a long comment;
Not sure how to proceed, but one way to rewrite it is $x=e^{iphi}$, getting
$$sum_{k=1}^n (2cos kphi)^k=sum_{k=1}^n 2^k T_k^k(cos(phi))$$
where $T$ is a Chebyshev polynomial of the first kind.
It seems to me that the expression is too convoluted to expect a nice solution. Another similar approach that might be considered is to express everything with $x+x^{-1}$. We have expressions of the type
$$(x+x^{-1})^2=2+(x^2+x^{-2})$$
$$(x+x^{-1})^3=3(x+x^{-1})+(x^3+x^{-3})$$
which could be solved for $x^k+x^{-k}$ but I can't see how to get a closed form for a general term. This solution does give you the constant term (reproduces result by @RobertIsrael) but the rest gets complicated due to $()^k$ - we get summations over very strange sets.
A reference that might be useful - it includes some very similar expressions.
answered Jan 18 at 14:26
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
Not an answer, just a long comment;
Not sure how to proceed, but one way to rewrite it is $x=e^{iphi}$, getting
$$sum_{k=1}^n (2cos kphi)^k=sum_{k=1}^n 2^k T_k^k(cos(phi))$$
where $T$ is a Chebyshev polynomial of the first kind.
It seems to me that the expression is too convoluted to expect a nice solution. Another similar approach that might be considered is to express everything with $x+x^{-1}$. We have expressions of the type
$$(x+x^{-1})^2=2+(x^2+x^{-2})$$
$$(x+x^{-1})^3=3(x+x^{-1})+(x^3+x^{-3})$$
which could be solved for $x^k+x^{-k}$ but I can't see how to get a closed form for a general term. This solution does give you the constant term (reproduces result by @RobertIsrael) but the rest gets complicated due to $()^k$ - we get summations over very strange sets.
A reference that might be useful - it includes some very similar expressions.
answered Jan 18 at 14:26
orionorion
13.6k11837
$endgroup$
add a comment |
$begingroup$
Not an answer, just a long comment;
Not sure how to proceed, but one way to rewrite it is $x=e^{iphi}$, getting
$$sum_{k=1}^n (2cos kphi)^k=sum_{k=1}^n 2^k T_k^k(cos(phi))$$
where $T$ is a Chebyshev polynomial of the first kind.
It seems to me that the expression is too convoluted to expect a nice solution. Another similar approach that might be considered is to express everything with $x+x^{-1}$. We have expressions of the type
$$(x+x^{-1})^2=2+(x^2+x^{-2})$$
$$(x+x^{-1})^3=3(x+x^{-1})+(x^3+x^{-3})$$
which could be solved for $x^k+x^{-k}$ but I can't see how to get a closed form for a general term. This solution does give you the constant term (reproduces result by @RobertIsrael) but the rest gets complicated due to $()^k$ - we get summations over very strange sets.
A reference that might be useful - it includes some very similar expressions.
answered Jan 18 at 14:26
orionorion
13.6k11837
$endgroup$
Not an answer, just a long comment;
Not sure how to proceed, but one way to rewrite it is $x=e^{iphi}$, getting
$$sum_{k=1}^n (2cos kphi)^k=sum_{k=1}^n 2^k T_k^k(cos(phi))$$
where $T$ is a Chebyshev polynomial of the first kind.
It seems to me that the expression is too convoluted to expect a nice solution. Another similar approach that might be considered is to express everything with $x+x^{-1}$. We have expressions of the type
$$(x+x^{-1})^2=2+(x^2+x^{-2})$$
$$(x+x^{-1})^3=3(x+x^{-1})+(x^3+x^{-3})$$
which could be solved for $x^k+x^{-k}$ but I can't see how to get a closed form for a general term. This solution does give you the constant term (reproduces result by @RobertIsrael) but the rest gets complicated due to $()^k$ - we get summations over very strange sets.
A reference that might be useful - it includes some very similar expressions.
answered Jan 18 at 14:26
orionorion
13.6k11837
answered Jan 18 at 14:26
orionorion
13.6k11837
answered Jan 18 at 14:26
orionorion
13.6k11837
answered Jan 18 at 14:26
orionorion
13.6k11837
13.6k11837
add a comment |
add a comment |
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$begingroup$
It's from an Olympiad paper I believe, so I assumed there would be a neat solution, with no need for double sums etc.
$endgroup$
– wrb98
Mar 8 '17 at 20:46
2
$begingroup$
I honestly have no clue. After binomial expansion, you end up with things like $x^{k^2}$, which have no closed form sum.
$endgroup$
– Simply Beautiful Art
Mar 8 '17 at 20:47
$begingroup$
Exactly. I tried to see if expanding the first brackets could lead to some pattern and then proceed by induction, but it's a dead end.
$endgroup$
– wrb98
Mar 8 '17 at 20:49
1
$begingroup$
Interesting, even Wolfram|Alpha can't figure it out.
$endgroup$
– projectilemotion
Mar 8 '17 at 20:52
3
$begingroup$
@Will: is the original problem really about simplifying such sum, or just about evaluating the coefficient of $x^0$ in the Laurent series?
$endgroup$
– Jack D'Aurizio
Mar 8 '17 at 21:06