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Math Help - Euler polynomials

  1. #1
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    Smile Euler polynomials

    Hello
    let B_m(x)=  \sum_{n=0}^m \frac{1}{n+1} \sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m
    prove that (B_0,B_1,...,B_m) is a basis of \mathbb{R}_m[X].
    thanks.
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  2. #2
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    Quote Originally Posted by Raoh View Post
    Hello
    let B_m(x)= \sum_{n=0}^m \frac{1}{n+1} \sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m
    prove that (B_0,B_1,...,B_m) is a basis of \mathbb{R}_m[X].
    thanks.
    they're called Bernoulli polynomials not Euler polynomials! anyway, we only need ro prove that B_0(x), \cdots , B_m(x) are \mathbb{R}-linearly independent because \dim_{\mathbb{R}} \mathbb{R}_m[x]=m+1.

    that can be easily proved by induction over m and using this fact that \frac{d}{dx} B_n(x)=nB_{n-1}(x), \ \forall n \in \mathbb{N}.
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  3. #3
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    Quote Originally Posted by NonCommAlg View Post
    they're called Bernoulli polynomials not Euler polynomials! anyway, we only need ro prove that B_0(x), \cdots , B_m(x) are \mathbb{R}-linearly independent because \dim_{\mathbb{R}} \mathbb{R}_m[x]=m+1.

    that can be easily proved by induction over m and using this fact that \frac{d}{dx} B_n(x)=nB_{n-1}(x), \ \forall n \in \mathbb{N}.
    i don't see how we can prove that they are linearly independent only by using the fact that \dim \mathbb{R}_m[x]=m+1..
    would you please explain a little bit.
    thanks.
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  4. #4
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    Quote Originally Posted by Raoh View Post
    i don't see how we can prove that they are linearly independent only by using the fact that \dim \mathbb{R}_m[x]=m+1..
    would you please explain a little bit.
    thanks.
    did you even read my post? i said since \dim \mathbb{R}_m[x]=m+1, we only need to prove that B_j(x), \ j=0, \cdots , m, are linearly independent. now, how can we prove that they are linearly independent? i suggested that you use induction over m and this fact that \frac{d}{dx}B_n(x)=nB_{n-1}(x). how? well, for m = 1, we have B_0(x)=1, \ B_1(x)=x - \frac{1}{2}, which are obviously linearly
    independent. now suppose the claim is true for m and \sum_{j=0}^{m+1} c_jB_j(x)=0, for some c_j \in \mathbb{R}. differentiating with respect to x will give us: 0=\sum_{j=0}^{m+1} c_j \frac{d}{dx} B_j(x)=\sum_{j=1}^{m+1}jc_j B_{j-1}(x)=\sum_{j=0}^m (j+1)c_{j+1}B_j(x). therefore, by the induction hypothesis, we'll have (j+1)c_{j+1}=0 and thus c_j=0, for all 0 \leq j \leq m, and hence c_{m+1}=0
    because \sum_{j=0}^{m+1} c_jB_j(x)=0.
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  5. #5
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    Quote Originally Posted by NonCommAlg View Post
    did you even read my post? i said since \dim \mathbb{R}_m[x]=m+1, we only need to prove that B_j(x), \ j=0, \cdots , m, are linearly independent. now, how can we prove that they are linearly independent? i suggested that you use induction over m and this fact that \frac{d}{dx}B_n(x)=nB_{n-1}(x). how? well, for m = 1, we have B_0(x)=1, \ B_1(x)=x - \frac{1}{2}, which are obviously linearly
    independent. now suppose the claim is true for m and \sum_{j=0}^{m+1} c_jB_j(x)=0, for some c_j \in \mathbb{R}. differentiating with respect to x will give us: 0=\sum_{j=0}^{m+1} c_j \frac{d}{dx} B_j(x)=\sum_{j=1}^{m+1}jc_j B_{j-1}(x)=\sum_{j=0}^m (j+1)c_{j+1}B_j(x). therefore, by the induction hypothesis, we'll have (j+1)c_{j+1}=0 and thus c_j=0, for all 0 \leq j \leq m, and hence c_{m+1}=0
    because \sum_{j=0}^{m+1} c_jB_j(x)=0.
    thank you
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