1. ## Euler polynomials

Hello
let $\displaystyle B_m(x)= \sum_{n=0}^m \frac{1}{n+1} \sum_{k=0}^n (-1)^k {n \choose k} (x+k)^m$
prove that $\displaystyle (B_0,B_1,...,B_m)$ is a basis of $\displaystyle \mathbb{R}_m[X]$.
thanks.

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

that can be easily proved by induction over $\displaystyle m$ and using this fact that $\displaystyle \frac{d}{dx} B_n(x)=nB_{n-1}(x), \ \forall n \in \mathbb{N}.$

3. Originally Posted by NonCommAlg
they're called Bernoulli polynomials not Euler polynomials! anyway, we only need ro prove that $\displaystyle B_0(x), \cdots , B_m(x)$ are $\displaystyle \mathbb{R}$-linearly independent because $\displaystyle \dim_{\mathbb{R}} \mathbb{R}_m[x]=m+1.$

that can be easily proved by induction over $\displaystyle m$ and using this fact that $\displaystyle \frac{d}{dx} B_n(x)=nB_{n-1}(x), \ \forall n \in \mathbb{N}.$
i don't see how we can prove that they are linearly independent only by using the fact that $\displaystyle \dim \mathbb{R}_m[x]=m+1.$.
would you please explain a little bit.
thanks.

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

5. Originally Posted by NonCommAlg
did you even read my post? i said since $\displaystyle \dim \mathbb{R}_m[x]=m+1,$ we only need to prove that $\displaystyle B_j(x), \ j=0, \cdots , m,$ are linearly independent. now, how can we prove that they are linearly independent? i suggested that you use induction over $\displaystyle m$ and this fact that $\displaystyle \frac{d}{dx}B_n(x)=nB_{n-1}(x).$ how? well, for m = 1, we have $\displaystyle B_0(x)=1, \ B_1(x)=x - \frac{1}{2},$ which are obviously linearly
independent. now suppose the claim is true for $\displaystyle m$ and $\displaystyle \sum_{j=0}^{m+1} c_jB_j(x)=0,$ for some $\displaystyle c_j \in \mathbb{R}.$ differentiating with respect to $\displaystyle x$ will give us: $\displaystyle 0=\sum_{j=0}^{m+1} c_j \frac{d}{dx} B_j(x)=\sum_{j=1}^{m+1}jc_j B_{j-1}(x)=\sum_{j=0}^m (j+1)c_{j+1}B_j(x).$ therefore, by the induction hypothesis, we'll have $\displaystyle (j+1)c_{j+1}=0$ and thus $\displaystyle c_j=0,$ for all $\displaystyle 0 \leq j \leq m,$ and hence $\displaystyle c_{m+1}=0$
because $\displaystyle \sum_{j=0}^{m+1} c_jB_j(x)=0.$
thank you