Binomial sum for an arbitrary function
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I'm looking for some known results for sum of this type but I can't find anything.
The sum is defined as:
$$S(x,a,b,n)=sum_{k=0}^n binom{n}{k} (-1)^{k} f((a(n-k)+bk)x)$$
where $f$ is an arbitrary function and $a$ and $b$ are real and positive constants.
For example for the function $f(x)=e^{x}$ the sum can be evaluated with Binomial Theorem and we have:
$$S(x,a,b,n)=(e^{ax}-e^{bx})^{n}$$
But for arbitrary function $f$?
Thanks.
functions summation binomial-coefficients integral-transforms
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
$endgroup$
add a comment |
$begingroup$
I'm looking for some known results for sum of this type but I can't find anything.
The sum is defined as:
$$S(x,a,b,n)=sum_{k=0}^n binom{n}{k} (-1)^{k} f((a(n-k)+bk)x)$$
where $f$ is an arbitrary function and $a$ and $b$ are real and positive constants.
For example for the function $f(x)=e^{x}$ the sum can be evaluated with Binomial Theorem and we have:
$$S(x,a,b,n)=(e^{ax}-e^{bx})^{n}$$
But for arbitrary function $f$?
Thanks.
functions summation binomial-coefficients integral-transforms
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
$endgroup$
$begingroup$
I suspect there is no compact representation of the sum without knowledge of $f$.
$endgroup$
– David G. Stork
Dec 19 '18 at 16:43
$begingroup$
There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:35
$begingroup$
Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results.
$endgroup$
– Papemax89
Dec 19 '18 at 17:40
add a comment |
$begingroup$
I'm looking for some known results for sum of this type but I can't find anything.
The sum is defined as:
$$S(x,a,b,n)=sum_{k=0}^n binom{n}{k} (-1)^{k} f((a(n-k)+bk)x)$$
where $f$ is an arbitrary function and $a$ and $b$ are real and positive constants.
For example for the function $f(x)=e^{x}$ the sum can be evaluated with Binomial Theorem and we have:
$$S(x,a,b,n)=(e^{ax}-e^{bx})^{n}$$
But for arbitrary function $f$?
Thanks.
functions summation binomial-coefficients integral-transforms
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
$endgroup$
I'm looking for some known results for sum of this type but I can't find anything.
The sum is defined as:
$$S(x,a,b,n)=sum_{k=0}^n binom{n}{k} (-1)^{k} f((a(n-k)+bk)x)$$
where $f$ is an arbitrary function and $a$ and $b$ are real and positive constants.
For example for the function $f(x)=e^{x}$ the sum can be evaluated with Binomial Theorem and we have:
$$S(x,a,b,n)=(e^{ax}-e^{bx})^{n}$$
But for arbitrary function $f$?
Thanks.
functions summation binomial-coefficients integral-transforms
functions summation binomial-coefficients integral-transforms
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
edited Dec 19 '18 at 16:58
Papemax89
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
asked Dec 19 '18 at 16:33
Papemax89Papemax89
162
162
$begingroup$
I suspect there is no compact representation of the sum without knowledge of $f$.
$endgroup$
– David G. Stork
Dec 19 '18 at 16:43
$begingroup$
There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:35
$begingroup$
Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results.
$endgroup$
– Papemax89
Dec 19 '18 at 17:40
add a comment |
$begingroup$
I suspect there is no compact representation of the sum without knowledge of $f$.
$endgroup$
– David G. Stork
Dec 19 '18 at 16:43
$begingroup$
There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:35
$begingroup$
Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results.
$endgroup$
– Papemax89
Dec 19 '18 at 17:40
$begingroup$
I suspect there is no compact representation of the sum without knowledge of $f$.
$endgroup$
– David G. Stork
Dec 19 '18 at 16:43
$begingroup$
I suspect there is no compact representation of the sum without knowledge of $f$.
$endgroup$
– David G. Stork
Dec 19 '18 at 16:43
$begingroup$
There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:35
$begingroup$
There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:35
$begingroup$
Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results.
$endgroup$
– Papemax89
Dec 19 '18 at 17:40
$begingroup$
Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results.
$endgroup$
– Papemax89
Dec 19 '18 at 17:40
add a comment |
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$begingroup$
I suspect there is no compact representation of the sum without knowledge of $f$.
$endgroup$
– David G. Stork
Dec 19 '18 at 16:43
$begingroup$
There probably isn't a single method for evaluating the sum that works for all functions. If you happen to know a little bit about $f$, you can make certain generalizations, though. For example, if $f$ is linear, then $f((a(n-k)+bk)x)=f(anx)-f(akx)+f(bkx)$.
$endgroup$
– R. Burton
Dec 19 '18 at 17:35
$begingroup$
Thanks!! Yes, for an arbitrary function isn't improbable to obtain a closed form. I tried to obtain an Integral representation of the sum via Integral Transform but with no great results.
$endgroup$
– Papemax89
Dec 19 '18 at 17:40