Constructing solutions for a given series to make it convergent
0
$begingroup$
I would like to find a function f such that $$y_i=f(i) geq 0$$ such that:
$$sum_{i=0}^n y_i = C(1-frac{3p}{4})^{n(1-frac{3p}{4})}(frac{3p}{4})^{n(frac{3p}{4})},$$
where $$C > 0, 0 leq p leq 1,$$ and
$$sum_{w=0}^nsum_{k=0}^nsum_{i=0}^n (-frac{1}{3})^kbinom{n-i}{w-k}binom{i}{k}3^{i}y_i$$ converges to a constant as $$nrightarrow infty.$$
Do you have any ideas? First of all, suggestions how to decompose the first sum would be useful.
Thanks.
sequences-and-series combinatorics limits convergence summation
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
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add a comment |
0
$begingroup$
I would like to find a function f such that $$y_i=f(i) geq 0$$ such that:
$$sum_{i=0}^n y_i = C(1-frac{3p}{4})^{n(1-frac{3p}{4})}(frac{3p}{4})^{n(frac{3p}{4})},$$
where $$C > 0, 0 leq p leq 1,$$ and
$$sum_{w=0}^nsum_{k=0}^nsum_{i=0}^n (-frac{1}{3})^kbinom{n-i}{w-k}binom{i}{k}3^{i}y_i$$ converges to a constant as $$nrightarrow infty.$$
Do you have any ideas? First of all, suggestions how to decompose the first sum would be useful.
Thanks.
sequences-and-series combinatorics limits convergence summation
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
$endgroup$
$begingroup$
What is $C$? If it is a constant independent on $n$ then for $C=1$ then $y_i=y_i(1)$ are determined uniquely, so we have only to check the convergence of the sum. For any other $C$ $y_i=y_i(C)=C(y_i(1))$, which doesn’t change the convergence of the sum.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 10:44
$begingroup$
Thanks for your comment. $C$ is a constant independent of $n$. However, I am not sure if I understand your notation. By $y_i=f(i)$ I mean that $y$'s are generated by some function of $i$. What do you mean by $y_i(1)$?
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:38
$begingroup$
By $y_i(1)$ I mean the (uniquely determined) value of $y_i$ when $C$ is $1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:40
$begingroup$
I see, yeah. My problem is to actually find which function may work here. I put $C$ to remember that I can scale without changing the convergence.
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:44
$begingroup$
If $p=0$ we encounter undefined expression $0^0$, but if $p>0$ then the function $f$ is defined uniquely by $y_0=C$ and $y_i=C(1-frac{3p}{4})^{i(1-frac{3p}{4})}(frac{3p}{4})^{i(frac{3p}{4})}- C(1-frac{3p}{4})^{(i-1)(1-frac{3p}{4})}(frac{3p}{4})^{(i-1)(frac{3p}{4})}$ for $i>1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:53
add a comment |
0
0
0
$begingroup$
I would like to find a function f such that $$y_i=f(i) geq 0$$ such that:
$$sum_{i=0}^n y_i = C(1-frac{3p}{4})^{n(1-frac{3p}{4})}(frac{3p}{4})^{n(frac{3p}{4})},$$
where $$C > 0, 0 leq p leq 1,$$ and
$$sum_{w=0}^nsum_{k=0}^nsum_{i=0}^n (-frac{1}{3})^kbinom{n-i}{w-k}binom{i}{k}3^{i}y_i$$ converges to a constant as $$nrightarrow infty.$$
Do you have any ideas? First of all, suggestions how to decompose the first sum would be useful.
Thanks.
sequences-and-series combinatorics limits convergence summation
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
$endgroup$
I would like to find a function f such that $$y_i=f(i) geq 0$$ such that:
$$sum_{i=0}^n y_i = C(1-frac{3p}{4})^{n(1-frac{3p}{4})}(frac{3p}{4})^{n(frac{3p}{4})},$$
where $$C > 0, 0 leq p leq 1,$$ and
$$sum_{w=0}^nsum_{k=0}^nsum_{i=0}^n (-frac{1}{3})^kbinom{n-i}{w-k}binom{i}{k}3^{i}y_i$$ converges to a constant as $$nrightarrow infty.$$
Do you have any ideas? First of all, suggestions how to decompose the first sum would be useful.
Thanks.
sequences-and-series combinatorics limits convergence summation
sequences-and-series combinatorics limits convergence summation
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
asked Dec 12 '18 at 19:54
GhostwriterGhostwriter
1
1
$begingroup$
What is $C$? If it is a constant independent on $n$ then for $C=1$ then $y_i=y_i(1)$ are determined uniquely, so we have only to check the convergence of the sum. For any other $C$ $y_i=y_i(C)=C(y_i(1))$, which doesn’t change the convergence of the sum.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 10:44
$begingroup$
Thanks for your comment. $C$ is a constant independent of $n$. However, I am not sure if I understand your notation. By $y_i=f(i)$ I mean that $y$'s are generated by some function of $i$. What do you mean by $y_i(1)$?
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:38
$begingroup$
By $y_i(1)$ I mean the (uniquely determined) value of $y_i$ when $C$ is $1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:40
$begingroup$
I see, yeah. My problem is to actually find which function may work here. I put $C$ to remember that I can scale without changing the convergence.
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:44
$begingroup$
If $p=0$ we encounter undefined expression $0^0$, but if $p>0$ then the function $f$ is defined uniquely by $y_0=C$ and $y_i=C(1-frac{3p}{4})^{i(1-frac{3p}{4})}(frac{3p}{4})^{i(frac{3p}{4})}- C(1-frac{3p}{4})^{(i-1)(1-frac{3p}{4})}(frac{3p}{4})^{(i-1)(frac{3p}{4})}$ for $i>1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:53
add a comment |
$begingroup$
What is $C$? If it is a constant independent on $n$ then for $C=1$ then $y_i=y_i(1)$ are determined uniquely, so we have only to check the convergence of the sum. For any other $C$ $y_i=y_i(C)=C(y_i(1))$, which doesn’t change the convergence of the sum.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 10:44
$begingroup$
Thanks for your comment. $C$ is a constant independent of $n$. However, I am not sure if I understand your notation. By $y_i=f(i)$ I mean that $y$'s are generated by some function of $i$. What do you mean by $y_i(1)$?
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:38
$begingroup$
By $y_i(1)$ I mean the (uniquely determined) value of $y_i$ when $C$ is $1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:40
$begingroup$
I see, yeah. My problem is to actually find which function may work here. I put $C$ to remember that I can scale without changing the convergence.
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:44
$begingroup$
If $p=0$ we encounter undefined expression $0^0$, but if $p>0$ then the function $f$ is defined uniquely by $y_0=C$ and $y_i=C(1-frac{3p}{4})^{i(1-frac{3p}{4})}(frac{3p}{4})^{i(frac{3p}{4})}- C(1-frac{3p}{4})^{(i-1)(1-frac{3p}{4})}(frac{3p}{4})^{(i-1)(frac{3p}{4})}$ for $i>1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:53
$begingroup$
What is $C$? If it is a constant independent on $n$ then for $C=1$ then $y_i=y_i(1)$ are determined uniquely, so we have only to check the convergence of the sum. For any other $C$ $y_i=y_i(C)=C(y_i(1))$, which doesn’t change the convergence of the sum.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 10:44
$begingroup$
What is $C$? If it is a constant independent on $n$ then for $C=1$ then $y_i=y_i(1)$ are determined uniquely, so we have only to check the convergence of the sum. For any other $C$ $y_i=y_i(C)=C(y_i(1))$, which doesn’t change the convergence of the sum.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 10:44
$begingroup$
Thanks for your comment. $C$ is a constant independent of $n$. However, I am not sure if I understand your notation. By $y_i=f(i)$ I mean that $y$'s are generated by some function of $i$. What do you mean by $y_i(1)$?
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:38
$begingroup$
Thanks for your comment. $C$ is a constant independent of $n$. However, I am not sure if I understand your notation. By $y_i=f(i)$ I mean that $y$'s are generated by some function of $i$. What do you mean by $y_i(1)$?
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:38
$begingroup$
By $y_i(1)$ I mean the (uniquely determined) value of $y_i$ when $C$ is $1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:40
$begingroup$
By $y_i(1)$ I mean the (uniquely determined) value of $y_i$ when $C$ is $1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:40
$begingroup$
I see, yeah. My problem is to actually find which function may work here. I put $C$ to remember that I can scale without changing the convergence.
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:44
$begingroup$
I see, yeah. My problem is to actually find which function may work here. I put $C$ to remember that I can scale without changing the convergence.
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:44
$begingroup$
If $p=0$ we encounter undefined expression $0^0$, but if $p>0$ then the function $f$ is defined uniquely by $y_0=C$ and $y_i=C(1-frac{3p}{4})^{i(1-frac{3p}{4})}(frac{3p}{4})^{i(frac{3p}{4})}- C(1-frac{3p}{4})^{(i-1)(1-frac{3p}{4})}(frac{3p}{4})^{(i-1)(frac{3p}{4})}$ for $i>1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:53
$begingroup$
If $p=0$ we encounter undefined expression $0^0$, but if $p>0$ then the function $f$ is defined uniquely by $y_0=C$ and $y_i=C(1-frac{3p}{4})^{i(1-frac{3p}{4})}(frac{3p}{4})^{i(frac{3p}{4})}- C(1-frac{3p}{4})^{(i-1)(1-frac{3p}{4})}(frac{3p}{4})^{(i-1)(frac{3p}{4})}$ for $i>1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:53
add a comment |
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$begingroup$
What is $C$? If it is a constant independent on $n$ then for $C=1$ then $y_i=y_i(1)$ are determined uniquely, so we have only to check the convergence of the sum. For any other $C$ $y_i=y_i(C)=C(y_i(1))$, which doesn’t change the convergence of the sum.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 10:44
$begingroup$
Thanks for your comment. $C$ is a constant independent of $n$. However, I am not sure if I understand your notation. By $y_i=f(i)$ I mean that $y$'s are generated by some function of $i$. What do you mean by $y_i(1)$?
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:38
$begingroup$
By $y_i(1)$ I mean the (uniquely determined) value of $y_i$ when $C$ is $1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:40
$begingroup$
I see, yeah. My problem is to actually find which function may work here. I put $C$ to remember that I can scale without changing the convergence.
$endgroup$
– Ghostwriter
Dec 15 '18 at 20:44
$begingroup$
If $p=0$ we encounter undefined expression $0^0$, but if $p>0$ then the function $f$ is defined uniquely by $y_0=C$ and $y_i=C(1-frac{3p}{4})^{i(1-frac{3p}{4})}(frac{3p}{4})^{i(frac{3p}{4})}- C(1-frac{3p}{4})^{(i-1)(1-frac{3p}{4})}(frac{3p}{4})^{(i-1)(frac{3p}{4})}$ for $i>1$.
$endgroup$
– Alex Ravsky
Dec 15 '18 at 20:53