[–]Number Theory 1 point2 points

Your approach is clever, but I think it's easier than you're making it. If you just do a change of variables v=m(u-x)+x, then we can set

[; B_n^m(v;x) = \frac{1}{m}B_n\left(\frac{v-x}{m}+x\right) ;]

and the substitution becomes

[; \int_x^{x+1}B_n(u)du = \int_x^{x+m} B_n^m(v;x) dv = x^n ;]

EDIT: I suppose the new polynomials depend on x, which is not very good

[–][S] 0 points1 point

I have never thought about going at it with substitution. This is great! It's true that the polynomials shouldn't depend on x, but maybe there's another substitution that gets rid of it.

[–]Number Theory 1 point2 points

I'm skeptical that a correct substitution will work. However, if you break up the integral from x to x+m of Bn(u) into (x,x+1), (x+1,x+2),...,(x+m-1,x+m) after integrating these piece by piece you get the sum from k=0 to m-1 of (x+k)n . You can then write this as a degree n polynomial where the ith coefficient is An,i,m:

[; \int_{x}^{x+m} B_n(u)du = \sum_{k=0}^{m-1}\int_{x+k}^{x+k+1}B_n(u)du = \sum_{k=0}^{m-1}(x+k)^n = \sum_{i=0}^{n} A_{n,i,m}x^i ;]

Explicitly, we have that

[; A_{n,i,m} = \binom{n}{i} \sum_{k=0}^{m-1} k^{n-i} ;]

which can also be written as:

[; A_{n,i,m} = \binom{n}{i}\frac{B_{n-i+1}(m)-B_{n-i+1}(0)}{n-i+1} ;]

Let's call result of the integral [; \mathcal{B}_{n,m}(x) ;]. So

[; \int_{x}^{x+m} B_n(u)du = \mathcal{B}_{n,m}(x) ;]

Notice that [; \mathcal{B}_{n,m}(x) ;] is a degree n polynomial and that the coefficient of xn is m. So we should be able to write [;\mathcal{B}_{n,m}(x);] as follows:

[; \mathcal{B}_{n,m}(x) = mx^n+\sum_{i=0}^{n-1} C_{n,i,m} \mathcal{B}_{i,m}(x) ;]

for some constants [;C_{n,i,m};] . This means that

[; x^n = \frac{ \mathcal{B}_{n,m}(x)-\sum_{i=0}^{n-1} C_{n,i,m} \mathcal{B}_{i,m}(x)}{m} ;]

so we should get

[; \int_{x}^{x+m} \frac{{B}_n(u)-\sum_{i=0}^{n-1} C_{n,i,m} B_{i}(u)}{m}du = x^n ;]

So the polynomials [; \mathfrak{B}_{n,m}(x)=\frac{{B}_n(x)-\sum_{i=0}^{n-1} C_{n,i,m} B_{i}(x)}{m};] have the desired property. By some uniqueness stuff that I'm not going to think about, this is probably the only polynomial with this property. Not the nicest thing, definitely doesn't have an immediately nice generating function like the regular Bernoulli polynomials, but it exists and depends only on n and m and nothing else.

EDIT: In fact, it looks like polynomials of different m can be written in terms of the divisors. Try and show that there are [; c_{n,j,m,k} ;] so that

[; \mathfrak{B}_{n,km}(x)=\frac{\mathfrak{B}_{n,m}(x)-\sum_{j=0}^{n-1} c_{n,j,m,k} \mathfrak{B}_{j,m}(x)}{k};]

I find that pretty interesting.
The text passage discusses a mathematical problem involving Bernoulli polynomials and their properties. It includes a detailed derivation and transformation of these polynomials using integrals and substitutions.
no_question

the passage does not contain a specific mathematical question or request for an answer. instead, it provides a detailed explanation and derivation of a mathematical concept.
Given a text passage, determine:
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    2. If an answer to that mathematical question is provided in the text

First, describe the content structure you found in 1-2 sentences.

Then provide your analysis in the following format:
TYPE: <one of the following values>
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system
[–]Number Theory 1 point2 points

Your approach is clever, but I think it's easier than you're making it. If you just do a change of variables v=m(u-x)+x, then we can set

[; B_n^m(v;x) = \frac{1}{m}B_n\left(\frac{v-x}{m}+x\right) ;]

and the substitution becomes

[; \int_x^{x+1}B_n(u)du = \int_x^{x+m} B_n^m(v;x) dv = x^n ;]

EDIT: I suppose the new polynomials depend on x, which is not very good

[–][S] 0 points1 point

I have never thought about going at it with substitution. This is great! It's true that the polynomials shouldn't depend on x, but maybe there's another substitution that gets rid of it.

[–]Number Theory 1 point2 points

I'm skeptical that a correct substitution will work. However, if you break up the integral from x to x+m of Bn(u) into (x,x+1), (x+1,x+2),...,(x+m-1,x+m) after integrating these piece by piece you get the sum from k=0 to m-1 of (x+k)n . You can then write this as a degree n polynomial where the ith coefficient is An,i,m:

[; \int_{x}^{x+m} B_n(u)du = \sum_{k=0}^{m-1}\int_{x+k}^{x+k+1}B_n(u)du = \sum_{k=0}^{m-1}(x+k)^n = \sum_{i=0}^{n} A_{n,i,m}x^i ;]

Explicitly, we have that

[; A_{n,i,m} = \binom{n}{i} \sum_{k=0}^{m-1} k^{n-i} ;]

which can also be written as:

[; A_{n,i,m} = \binom{n}{i}\frac{B_{n-i+1}(m)-B_{n-i+1}(0)}{n-i+1} ;]

Let's call result of the integral [; \mathcal{B}_{n,m}(x) ;]. So

[; \int_{x}^{x+m} B_n(u)du = \mathcal{B}_{n,m}(x) ;]

Notice that [; \mathcal{B}_{n,m}(x) ;] is a degree n polynomial and that the coefficient of xn is m. So we should be able to write [;\mathcal{B}_{n,m}(x);] as follows:

[; \mathcal{B}_{n,m}(x) = mx^n+\sum_{i=0}^{n-1} C_{n,i,m} \mathcal{B}_{i,m}(x) ;]

for some constants [;C_{n,i,m};] . This means that

[; x^n = \frac{ \mathcal{B}_{n,m}(x)-\sum_{i=0}^{n-1} C_{n,i,m} \mathcal{B}_{i,m}(x)}{m} ;]

so we should get

[; \int_{x}^{x+m} \frac{{B}_n(u)-\sum_{i=0}^{n-1} C_{n,i,m} B_{i}(u)}{m}du = x^n ;]

So the polynomials [; \mathfrak{B}_{n,m}(x)=\frac{{B}_n(x)-\sum_{i=0}^{n-1} C_{n,i,m} B_{i}(x)}{m};] have the desired property. By some uniqueness stuff that I'm not going to think about, this is probably the only polynomial with this property. Not the nicest thing, definitely doesn't have an immediately nice generating function like the regular Bernoulli polynomials, but it exists and depends only on n and m and nothing else.

EDIT: In fact, it looks like polynomials of different m can be written in terms of the divisors. Try and show that there are [; c_{n,j,m,k} ;] so that

[; \mathfrak{B}_{n,km}(x)=\frac{\mathfrak{B}_{n,m}(x)-\sum_{j=0}^{n-1} c_{n,j,m,k} \mathfrak{B}_{j,m}(x)}{k};]

I find that pretty interesting.
user
The text passage discusses a mathematical problem involving Bernoulli polynomials and their properties. It includes a detailed derivation and transformation of these polynomials using integrals and substitutions.

TYPE: NO_QUESTION

The passage does not contain a specific mathematical question or request for an answer. Instead, it provides a detailed explanation and derivation of a mathematical concept.