text: [–]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.
explanation: The text extract is a discussion between users on a platform, likely Reddit, focusing on a mathematical problem involving Bernoulli polynomials and integrals. The discussion includes a detailed explanation of a substitution method and the derivation of a polynomial with specific properties. The text contains a series of mathematical steps and insights, but it does not present a clear, single question or a question-and-answer format. Instead, it is a collaborative exploration of a mathematical problem.
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distilabel_metadata.content: [–]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.
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distilabel_metadata.raw_output_q_a_detector_0: The text extract is a discussion between users on a platform, likely Reddit, focusing on a mathematical problem involving Bernoulli polynomials and integrals. The discussion includes a detailed explanation of a substitution method and the derivation of a polynomial with specific properties. The text contains a series of mathematical steps and insights, but it does not present a clear, single question or a question-and-answer format. Instead, it is a collaborative exploration of a mathematical problem.
  
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