Mixing
Equating the coefficients in $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!}...
$begingroup$
$$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}$$
In the above equation is it possible to equate the coefficients of $frac{t^k}{k!}$? Or do we need to consider $(frac{t^k}{k!})^2$? Could you please give me an idea for this?
sequences-and-series number-theory power-series
edited Jan 15 at 19:41
user
4,52011029
asked Jan 14 at 21:52
Soma WickSoma Wick
305
$endgroup$
add a comment |
$begingroup$
$$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}$$
In the above equation is it possible to equate the coefficients of $frac{t^k}{k!}$? Or do we need to consider $(frac{t^k}{k!})^2$? Could you please give me an idea for this?
sequences-and-series number-theory power-series
edited Jan 15 at 19:41
user
4,52011029
asked Jan 14 at 21:52
Soma WickSoma Wick
305
$endgroup$
$begingroup$
Please do not change the text of the question after you have received an answer. I would propose to rollback to the version 2.
$endgroup$
– user
Jan 15 at 16:30
$begingroup$
@user I understand. My main question remains the same. I couldn`t write this in the comment box. That was why I wrote the steps for you to show how I did this.
$endgroup$
– Soma Wick
Jan 15 at 17:46
add a comment |
$begingroup$
$$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}$$
In the above equation is it possible to equate the coefficients of $frac{t^k}{k!}$? Or do we need to consider $(frac{t^k}{k!})^2$? Could you please give me an idea for this?
sequences-and-series number-theory power-series
edited Jan 15 at 19:41
user
4,52011029
asked Jan 14 at 21:52
Soma WickSoma Wick
305
$endgroup$
$$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}$$
In the above equation is it possible to equate the coefficients of $frac{t^k}{k!}$? Or do we need to consider $(frac{t^k}{k!})^2$? Could you please give me an idea for this?
sequences-and-series number-theory power-series
sequences-and-series number-theory power-series
edited Jan 15 at 19:41
user
4,52011029
asked Jan 14 at 21:52
Soma WickSoma Wick
305
edited Jan 15 at 19:41
user
4,52011029
asked Jan 14 at 21:52
Soma WickSoma Wick
305
edited Jan 15 at 19:41
user
4,52011029
edited Jan 15 at 19:41
user
4,52011029
edited Jan 15 at 19:41
user
4,52011029
4,52011029
asked Jan 14 at 21:52
Soma WickSoma Wick
305
asked Jan 14 at 21:52
Soma WickSoma Wick
305
asked Jan 14 at 21:52
Soma WickSoma Wick
305
305
$begingroup$
Please do not change the text of the question after you have received an answer. I would propose to rollback to the version 2.
$endgroup$
– user
Jan 15 at 16:30
$begingroup$
@user I understand. My main question remains the same. I couldn`t write this in the comment box. That was why I wrote the steps for you to show how I did this.
$endgroup$
– Soma Wick
Jan 15 at 17:46
add a comment |
$begingroup$
Please do not change the text of the question after you have received an answer. I would propose to rollback to the version 2.
$endgroup$
– user
Jan 15 at 16:30
$begingroup$
@user I understand. My main question remains the same. I couldn`t write this in the comment box. That was why I wrote the steps for you to show how I did this.
$endgroup$
– Soma Wick
Jan 15 at 17:46
$begingroup$
Please do not change the text of the question after you have received an answer. I would propose to rollback to the version 2.
$endgroup$
– user
Jan 15 at 16:30
$begingroup$
Please do not change the text of the question after you have received an answer. I would propose to rollback to the version 2.
$endgroup$
– user
Jan 15 at 16:30
$begingroup$
@user I understand. My main question remains the same. I couldn`t write this in the comment box. That was why I wrote the steps for you to show how I did this.
$endgroup$
– Soma Wick
Jan 15 at 17:46
$begingroup$
@user I understand. My main question remains the same. I couldn`t write this in the comment box. That was why I wrote the steps for you to show how I did this.
$endgroup$
– Soma Wick
Jan 15 at 17:46
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Equating the coefficients at equal powers of $t$ on both sides one obtains:
$$C_{n-1}=sum_{k=0}^nfrac{n!}{k!(n-k)!}C_k.$$
The equating process goes as follows:
$$sum_{k=0}^infty C_k frac{t^k}{k!} sum_{m=0}^infty frac{t^m}{m!}
=sum_{k=0}^infty C_kfrac1{k!}sum_{m=0}^infty frac{t^{k+m}}{m!}\
stackrel{n=k+m}=sum_{k=0}^infty C_kfrac1{k!} sum_{n=k}^{infty} frac{t^{n}}{(n-k)!}
=sum_{n=0}^infty t^nsum_{k=0}^n C_k frac{1}{k!(n-k)!}\
=sum_{n=0}^inftyfrac{t^n}{n!}sum_{k=0}^n C_k frac{n!}{k!(n-k)!}.$$
The reason for the possibility of the exchange of the summation order in the second line is that both double sums run over the same set of indices: $0le kle n<infty$. It can be seen also from the fact that the summation in both $sum_{n=k}^{infty}$ and $sum_{k=0}^n$ formally can run from $0$ to infinity due to $(n-k)!$ term.
answered Jan 14 at 22:02
useruser
4,52011029
$endgroup$
1
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
1
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
1
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
2
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
1
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
|
show 5 more comments
$begingroup$
$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}
=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
$
Here's how you get
the binomial coefficients.
Note:
These are called exponential generating functions
because of their similarity to
$e^t
=sum_{m=0}^infty frac{t^m}{m!}
$.
$begin{array}\
sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
&=sum_{n=0}^inftysum_{m=0}^infty C_n frac{t^{n+m}}{n!m!}\
&=sum_{k=0}^inftysum_{n=0}^k C_n frac{t^{k}}{n!(k-n)!}
qquad n+m = k, m = k-n\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n frac{k!}{n!(k-n)!}\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n binom{k}{n}\
end{array}
$
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
$endgroup$
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
$begingroup$
Equating the coefficients at equal powers of $t$ on both sides one obtains:
$$C_{n-1}=sum_{k=0}^nfrac{n!}{k!(n-k)!}C_k.$$
The equating process goes as follows:
$$sum_{k=0}^infty C_k frac{t^k}{k!} sum_{m=0}^infty frac{t^m}{m!}
=sum_{k=0}^infty C_kfrac1{k!}sum_{m=0}^infty frac{t^{k+m}}{m!}\
stackrel{n=k+m}=sum_{k=0}^infty C_kfrac1{k!} sum_{n=k}^{infty} frac{t^{n}}{(n-k)!}
=sum_{n=0}^infty t^nsum_{k=0}^n C_k frac{1}{k!(n-k)!}\
=sum_{n=0}^inftyfrac{t^n}{n!}sum_{k=0}^n C_k frac{n!}{k!(n-k)!}.$$
The reason for the possibility of the exchange of the summation order in the second line is that both double sums run over the same set of indices: $0le kle n<infty$. It can be seen also from the fact that the summation in both $sum_{n=k}^{infty}$ and $sum_{k=0}^n$ formally can run from $0$ to infinity due to $(n-k)!$ term.
answered Jan 14 at 22:02
useruser
4,52011029
$endgroup$
1
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
1
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
1
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
2
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
1
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
|
show 5 more comments
$begingroup$
Equating the coefficients at equal powers of $t$ on both sides one obtains:
$$C_{n-1}=sum_{k=0}^nfrac{n!}{k!(n-k)!}C_k.$$
The equating process goes as follows:
$$sum_{k=0}^infty C_k frac{t^k}{k!} sum_{m=0}^infty frac{t^m}{m!}
=sum_{k=0}^infty C_kfrac1{k!}sum_{m=0}^infty frac{t^{k+m}}{m!}\
stackrel{n=k+m}=sum_{k=0}^infty C_kfrac1{k!} sum_{n=k}^{infty} frac{t^{n}}{(n-k)!}
=sum_{n=0}^infty t^nsum_{k=0}^n C_k frac{1}{k!(n-k)!}\
=sum_{n=0}^inftyfrac{t^n}{n!}sum_{k=0}^n C_k frac{n!}{k!(n-k)!}.$$
The reason for the possibility of the exchange of the summation order in the second line is that both double sums run over the same set of indices: $0le kle n<infty$. It can be seen also from the fact that the summation in both $sum_{n=k}^{infty}$ and $sum_{k=0}^n$ formally can run from $0$ to infinity due to $(n-k)!$ term.
answered Jan 14 at 22:02
useruser
4,52011029
$endgroup$
1
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
1
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
1
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
2
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
1
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
|
show 5 more comments
$begingroup$
Equating the coefficients at equal powers of $t$ on both sides one obtains:
$$C_{n-1}=sum_{k=0}^nfrac{n!}{k!(n-k)!}C_k.$$
The equating process goes as follows:
$$sum_{k=0}^infty C_k frac{t^k}{k!} sum_{m=0}^infty frac{t^m}{m!}
=sum_{k=0}^infty C_kfrac1{k!}sum_{m=0}^infty frac{t^{k+m}}{m!}\
stackrel{n=k+m}=sum_{k=0}^infty C_kfrac1{k!} sum_{n=k}^{infty} frac{t^{n}}{(n-k)!}
=sum_{n=0}^infty t^nsum_{k=0}^n C_k frac{1}{k!(n-k)!}\
=sum_{n=0}^inftyfrac{t^n}{n!}sum_{k=0}^n C_k frac{n!}{k!(n-k)!}.$$
The reason for the possibility of the exchange of the summation order in the second line is that both double sums run over the same set of indices: $0le kle n<infty$. It can be seen also from the fact that the summation in both $sum_{n=k}^{infty}$ and $sum_{k=0}^n$ formally can run from $0$ to infinity due to $(n-k)!$ term.
answered Jan 14 at 22:02
useruser
4,52011029
$endgroup$
Equating the coefficients at equal powers of $t$ on both sides one obtains:
$$C_{n-1}=sum_{k=0}^nfrac{n!}{k!(n-k)!}C_k.$$
The equating process goes as follows:
$$sum_{k=0}^infty C_k frac{t^k}{k!} sum_{m=0}^infty frac{t^m}{m!}
=sum_{k=0}^infty C_kfrac1{k!}sum_{m=0}^infty frac{t^{k+m}}{m!}\
stackrel{n=k+m}=sum_{k=0}^infty C_kfrac1{k!} sum_{n=k}^{infty} frac{t^{n}}{(n-k)!}
=sum_{n=0}^infty t^nsum_{k=0}^n C_k frac{1}{k!(n-k)!}\
=sum_{n=0}^inftyfrac{t^n}{n!}sum_{k=0}^n C_k frac{n!}{k!(n-k)!}.$$
The reason for the possibility of the exchange of the summation order in the second line is that both double sums run over the same set of indices: $0le kle n<infty$. It can be seen also from the fact that the summation in both $sum_{n=k}^{infty}$ and $sum_{k=0}^n$ formally can run from $0$ to infinity due to $(n-k)!$ term.
answered Jan 14 at 22:02
useruser
4,52011029
edited Jan 15 at 16:03
answered Jan 14 at 22:02
useruser
4,52011029
answered Jan 14 at 22:02
useruser
4,52011029
answered Jan 14 at 22:02
useruser
4,52011029
4,52011029
1
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
1
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
1
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
2
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
1
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
|
show 5 more comments
1
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
1
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
1
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
2
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
1
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
1
1
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
$begingroup$
Thank you for the help!
$endgroup$
– Soma Wick
Jan 14 at 22:15
1
1
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
$begingroup$
You're welcome.
$endgroup$
– user
Jan 14 at 22:16
1
1
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
$begingroup$
If we change the equation as $sum_{n=1}^infty C_{n-1} frac{t^n}{n!}=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^{2m+1}}{(2m+1)!}$, will the result be the same?
$endgroup$
– Soma Wick
Jan 15 at 0:28
2
2
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
$begingroup$
@SomaWick The result will be similar. The sum will be only over $k$ with parity different from that of $n$.
$endgroup$
– user
Jan 15 at 1:20
1
1
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
$begingroup$
Thank you. But I don`t understand how to get the binomial coefficient part. Could you please help me to understand it?
$endgroup$
– Soma Wick
Jan 15 at 2:21
|
show 5 more comments
$begingroup$
$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}
=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
$
Here's how you get
the binomial coefficients.
Note:
These are called exponential generating functions
because of their similarity to
$e^t
=sum_{m=0}^infty frac{t^m}{m!}
$.
$begin{array}\
sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
&=sum_{n=0}^inftysum_{m=0}^infty C_n frac{t^{n+m}}{n!m!}\
&=sum_{k=0}^inftysum_{n=0}^k C_n frac{t^{k}}{n!(k-n)!}
qquad n+m = k, m = k-n\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n frac{k!}{n!(k-n)!}\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n binom{k}{n}\
end{array}
$
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
$endgroup$
add a comment |
$begingroup$
$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}
=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
$
Here's how you get
the binomial coefficients.
Note:
These are called exponential generating functions
because of their similarity to
$e^t
=sum_{m=0}^infty frac{t^m}{m!}
$.
$begin{array}\
sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
&=sum_{n=0}^inftysum_{m=0}^infty C_n frac{t^{n+m}}{n!m!}\
&=sum_{k=0}^inftysum_{n=0}^k C_n frac{t^{k}}{n!(k-n)!}
qquad n+m = k, m = k-n\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n frac{k!}{n!(k-n)!}\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n binom{k}{n}\
end{array}
$
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
$endgroup$
add a comment |
$begingroup$
$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}
=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
$
Here's how you get
the binomial coefficients.
Note:
These are called exponential generating functions
because of their similarity to
$e^t
=sum_{m=0}^infty frac{t^m}{m!}
$.
$begin{array}\
sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
&=sum_{n=0}^inftysum_{m=0}^infty C_n frac{t^{n+m}}{n!m!}\
&=sum_{k=0}^inftysum_{n=0}^k C_n frac{t^{k}}{n!(k-n)!}
qquad n+m = k, m = k-n\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n frac{k!}{n!(k-n)!}\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n binom{k}{n}\
end{array}
$
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
$endgroup$
$sum_{n=1}^infty C_{n-1} frac{t^n}{n!}
=sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
$
Here's how you get
the binomial coefficients.
Note:
These are called exponential generating functions
because of their similarity to
$e^t
=sum_{m=0}^infty frac{t^m}{m!}
$.
$begin{array}\
sum_{n=0}^infty C_n frac{t^n}{n!} sum_{m=0}^infty frac{t^m}{m!}
&=sum_{n=0}^inftysum_{m=0}^infty C_n frac{t^{n+m}}{n!m!}\
&=sum_{k=0}^inftysum_{n=0}^k C_n frac{t^{k}}{n!(k-n)!}
qquad n+m = k, m = k-n\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n frac{k!}{n!(k-n)!}\
&=sum_{k=0}^inftyfrac{t^k}{k!}sum_{n=0}^k C_n binom{k}{n}\
end{array}
$
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
answered Jan 15 at 2:46
marty cohenmarty cohen
73.7k549128
73.7k549128
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$begingroup$
Please do not change the text of the question after you have received an answer. I would propose to rollback to the version 2.
$endgroup$
– user
Jan 15 at 16:30
$begingroup$
@user I understand. My main question remains the same. I couldn`t write this in the comment box. That was why I wrote the steps for you to show how I did this.
$endgroup$
– Soma Wick
Jan 15 at 17:46