Property of the Bernoulli PolynomialsBernoulli polynomials, ApostolRoots of some modified Bernoulli polynomialsProving $int_0^1 B_n(x) dx=0$ for Bernoulli polynomials$frac-12$ Zero of odd powers sum polynomials?Simplification of recursive polynomialsBernoulli Polynomials from Apostol's calculus bookMisunderstanding about Bernoulli-Euler relation?Multivariate Bell polynomialsidentity involving translation for Bernoulli's polynomialGenerating Function and Laguerre Polynomials.
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Property of the Bernoulli Polynomials
Bernoulli polynomials, ApostolRoots of some modified Bernoulli polynomialsProving $int_0^1 B_n(x) dx=0$ for Bernoulli polynomials$frac-12$ Zero of odd powers sum polynomials?Simplification of recursive polynomialsBernoulli Polynomials from Apostol's calculus bookMisunderstanding about Bernoulli-Euler relation?Multivariate Bell polynomialsidentity involving translation for Bernoulli's polynomialGenerating Function and Laguerre Polynomials.
$begingroup$
I'm trying to prove the following equality involving the Bernoulli polynomials:
$B_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright)$ for all $NinmathbbN$.
Since this is to be proved for all natural numbers, I immediately think that I need to use induction. Clearly the base case of $N=1$ is trivial. However, I'm not sure how to handle the general case. There are two ways that I've seen the Bernoulli polynomials presented:
A generating function: $fracte^txe^t-1=sum_k=0^inftyB_k(x)fract^kk!.$
Using the Bernoulli numbers: we define $B_k(x)=sum_n=0^kbinomknb_k-nx^n$, where $b_k-n$ are the Bernoulli numbers.
I know that there are many other ways to define the Bernoulli polynomials. At any rate, would either of the definitions above be easily adopted to prove the desired equality?
I tried a bit already with the second definition by substituting $B_kleft(fracx+aNright)=sum_n=0^kbinomknb_k-nleft(fracx+aNright)^n$, but I quickly started to trip over notation. Thanks in advance for any suggestions.
abstract-algebra polynomials bernoulli-polynomials
asked Mar 12 at 20:16
user608571user608571
25613
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the following equality involving the Bernoulli polynomials:
$B_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright)$ for all $NinmathbbN$.
Since this is to be proved for all natural numbers, I immediately think that I need to use induction. Clearly the base case of $N=1$ is trivial. However, I'm not sure how to handle the general case. There are two ways that I've seen the Bernoulli polynomials presented:
A generating function: $fracte^txe^t-1=sum_k=0^inftyB_k(x)fract^kk!.$
Using the Bernoulli numbers: we define $B_k(x)=sum_n=0^kbinomknb_k-nx^n$, where $b_k-n$ are the Bernoulli numbers.
I know that there are many other ways to define the Bernoulli polynomials. At any rate, would either of the definitions above be easily adopted to prove the desired equality?
I tried a bit already with the second definition by substituting $B_kleft(fracx+aNright)=sum_n=0^kbinomknb_k-nleft(fracx+aNright)^n$, but I quickly started to trip over notation. Thanks in advance for any suggestions.
abstract-algebra polynomials bernoulli-polynomials
asked Mar 12 at 20:16
user608571user608571
25613
$endgroup$
add a comment |
$begingroup$
I'm trying to prove the following equality involving the Bernoulli polynomials:
$B_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright)$ for all $NinmathbbN$.
Since this is to be proved for all natural numbers, I immediately think that I need to use induction. Clearly the base case of $N=1$ is trivial. However, I'm not sure how to handle the general case. There are two ways that I've seen the Bernoulli polynomials presented:
A generating function: $fracte^txe^t-1=sum_k=0^inftyB_k(x)fract^kk!.$
Using the Bernoulli numbers: we define $B_k(x)=sum_n=0^kbinomknb_k-nx^n$, where $b_k-n$ are the Bernoulli numbers.
I know that there are many other ways to define the Bernoulli polynomials. At any rate, would either of the definitions above be easily adopted to prove the desired equality?
I tried a bit already with the second definition by substituting $B_kleft(fracx+aNright)=sum_n=0^kbinomknb_k-nleft(fracx+aNright)^n$, but I quickly started to trip over notation. Thanks in advance for any suggestions.
abstract-algebra polynomials bernoulli-polynomials
asked Mar 12 at 20:16
user608571user608571
25613
$endgroup$
I'm trying to prove the following equality involving the Bernoulli polynomials:
$B_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright)$ for all $NinmathbbN$.
Since this is to be proved for all natural numbers, I immediately think that I need to use induction. Clearly the base case of $N=1$ is trivial. However, I'm not sure how to handle the general case. There are two ways that I've seen the Bernoulli polynomials presented:
A generating function: $fracte^txe^t-1=sum_k=0^inftyB_k(x)fract^kk!.$
Using the Bernoulli numbers: we define $B_k(x)=sum_n=0^kbinomknb_k-nx^n$, where $b_k-n$ are the Bernoulli numbers.
I know that there are many other ways to define the Bernoulli polynomials. At any rate, would either of the definitions above be easily adopted to prove the desired equality?
I tried a bit already with the second definition by substituting $B_kleft(fracx+aNright)=sum_n=0^kbinomknb_k-nleft(fracx+aNright)^n$, but I quickly started to trip over notation. Thanks in advance for any suggestions.
abstract-algebra polynomials bernoulli-polynomials
abstract-algebra polynomials bernoulli-polynomials
asked Mar 12 at 20:16
user608571user608571
25613
asked Mar 12 at 20:16
user608571user608571
25613
asked Mar 12 at 20:16
user608571user608571
25613
asked Mar 12 at 20:16
user608571user608571
25613
asked Mar 12 at 20:16
user608571user608571
25613
25613
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Write
$$C_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright).$$
Then
beginalign
sum_k=0^infty C_k(x)fract^kk!
&=frac1Nsum_a=0^N-1sum_k=0^infty B_kleft(fracx+aNright)
frac(Nt)^kk!
=frac1Nsum_a=0^N-1fracNte^t(x+a)e^Nt-1\
&=frace^txe^Nt-1sum_a=0^N-1e^at=frace^txe^t-1
=sum_k=0^infty B_k(x)fract^kk!.
endalign
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Write
$$C_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright).$$
Then
beginalign
sum_k=0^infty C_k(x)fract^kk!
&=frac1Nsum_a=0^N-1sum_k=0^infty B_kleft(fracx+aNright)
frac(Nt)^kk!
=frac1Nsum_a=0^N-1fracNte^t(x+a)e^Nt-1\
&=frace^txe^Nt-1sum_a=0^N-1e^at=frace^txe^t-1
=sum_k=0^infty B_k(x)fract^kk!.
endalign
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
add a comment |
$begingroup$
Write
$$C_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright).$$
Then
beginalign
sum_k=0^infty C_k(x)fract^kk!
&=frac1Nsum_a=0^N-1sum_k=0^infty B_kleft(fracx+aNright)
frac(Nt)^kk!
=frac1Nsum_a=0^N-1fracNte^t(x+a)e^Nt-1\
&=frace^txe^Nt-1sum_a=0^N-1e^at=frace^txe^t-1
=sum_k=0^infty B_k(x)fract^kk!.
endalign
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
add a comment |
$begingroup$
Write
$$C_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright).$$
Then
beginalign
sum_k=0^infty C_k(x)fract^kk!
&=frac1Nsum_a=0^N-1sum_k=0^infty B_kleft(fracx+aNright)
frac(Nt)^kk!
=frac1Nsum_a=0^N-1fracNte^t(x+a)e^Nt-1\
&=frace^txe^Nt-1sum_a=0^N-1e^at=frace^txe^t-1
=sum_k=0^infty B_k(x)fract^kk!.
endalign
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
$endgroup$
Write
$$C_k(x)=N^k-1sum_a=0^N-1B_kleft(fracx+aNright).$$
Then
beginalign
sum_k=0^infty C_k(x)fract^kk!
&=frac1Nsum_a=0^N-1sum_k=0^infty B_kleft(fracx+aNright)
frac(Nt)^kk!
=frac1Nsum_a=0^N-1fracNte^t(x+a)e^Nt-1\
&=frace^txe^Nt-1sum_a=0^N-1e^at=frace^txe^t-1
=sum_k=0^infty B_k(x)fract^kk!.
endalign
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
answered Mar 12 at 21:13
Lord Shark the UnknownLord Shark the Unknown
106k1161133
106k1161133
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
add a comment |
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
$begingroup$
This makes perfect sense now that I see it. Thank you!
$endgroup$
– user608571
Mar 12 at 21:29
add a comment |
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