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Bernstein polynomials are defined like this $B_i^n={{n}\choose{i}}x^i(1-x)^{n-i}$.I need to prove that $r$'th derivative of it is equal to:

$(B_i^n)^{(r)}=\dfrac{n!}{(n-r)!}\sum\limits_{k=0}^{r}(-1)^k{{r}\choose{k}}B_{i-r+k}^{n-r}=\dfrac{n!}{(n-r)!}\sum\limits_{k=0}^{r}(-1)^k{{r}\choose{k}}{{n-r}\choose{i-r+k}}x^{i-r+k}(1-x)^{n-i-k}$.

Here's my partial solution end eventual getting stuck:

$(B_i^n)^{(r)}=({{n}\choose{i}}x^i(1-x)^{n-i})^{(r)} = {{n}\choose{i}}(x^i(1-x)^{n-i})^{(r)} \stackrel{Leibniz}{=} {{n}\choose{i}}\sum\limits_{k=0}^{r}{{r}\choose{k}}(x^i)^{(k)}((1-x)^{n-i})^{(r-k)}={{n}\choose{i}}\sum\limits_{k=0}^{r}{{r}\choose{k}}\dfrac{i!}{(i-k)!}x^{i-k}\dfrac{(n-i)!}{(n-i-r+k)!}(1-x)^{n-i-r+k}(-1)^k $

What am I doing wrong?

Arek Krawczyk
  • 2,114

2 Answers2

1

You are on the right track. Your derivation is correct, but you need to to do more work to reach your destiny. Your last step

$$ {{n}\choose{i}}\sum\limits_{k=0}^{r}{{r}\choose{k}}\dfrac{i!}{(i-k)!}x^{i-k}\dfrac{(n-i)!}{(n-i-r+k)!}(1-x)^{n-i-r+k}(-1)^k $$

can be simplified as

$$ = \frac{n!}{(n-r)!}\sum\limits_{k=0}^{r}{{r}\choose{k}}\dfrac{1}{(i-k)!}\dfrac{(n-r)!}{(n-i-r+k)!}x^{i-k}(1-x)^{n-i-r+k}(-1)^k. $$

Now, see this

$$ \dfrac{(n-r)!}{(i-k)!(n-i-r+k)!} = { n-r \choose i-r+k } . $$

Note: You need to expand ${{n}\choose{i}}$ and simplify in your last step.

Mhenni Benghorbal
  • 47,431
1

Here is another contender that does not use the Leibniz rule. The term that needs to be simplified here is $$\frac{n!}{(n-r)!} \sum_{k=0}^r {r\choose k} (-1)^k {n-r\choose i-r+k} x^{i-r+k} (1-x)^{n-i-k}$$ which is $$\frac{n!}{(n-r)!} x^i (1-x)^{n-i} \sum_{k=0}^r {r\choose k} (-1)^k \frac{(n-r)!}{(i-(r-k))!(n-i-k)!} \frac{1}{x^{r-k}} \frac{1}{(1-x)^k}$$ which gives $$n! x^i (1-x)^{n-i} \sum_{k=0}^r {r\choose k} (-1)^k \frac{1}{(i-(r-k))!(n-i-k)!} \frac{1}{x^{r-k}} \frac{1}{(1-x)^k}$$ To evaluate the sum, observe that when we multiply two exponential generating functions of the sequences $\{a_n\}$ and $\{b_n\}$ we get that $$ A(z) B(z) = \sum_{n\ge 0} a_n \frac{z^n}{n!} \sum_{n\ge 0} b_n \frac{z^n}{n!} = \sum_{n\ge 0} \sum_{k=0}^n \frac{1}{k!}\frac{1}{(n-k)!} a_k b_{n-k} z^n\\ = \sum_{n\ge 0} \sum_{k=0}^n \frac{n!}{k!(n-k)!} a_k b_{n-k} \frac{z^n}{n!} = \sum_{n\ge 0} \left(\sum_{k=0}^n {n\choose k} a_k b_{n-k}\right)\frac{z^n}{n!}$$ i.e. the product of the two generating functions is the generating function of $$\sum_{k=0}^n {n\choose k} a_k b_{n-k}.$$

Now in the present case we clearly have $$A(z) = \sum_{q=0}^{n-i} \frac{(-1)^q}{(n-i-q)!} \frac{1}{(1-x)^q} \frac{z^q}{q!} = \frac{1}{(n-i)!} \sum_{q=0}^{n-i} {n-i\choose q} (-1)^q \frac{1}{(1-x)^q} z^q\\ = \frac{1}{(n-i)!} \left(1- \frac{z}{1-x}\right)^{n-i} = \frac{1}{(n-i)!} \left(\frac{1-x-z}{1-x}\right)^{n-i}.$$ For $B(z)$ we get that $$B(z) = \sum_{q=0}^i \frac{1}{(i-q)!} \frac{1}{x^q}\frac{z^q}{q!} = \frac{1}{i!} \sum_{q=0}^i {i\choose q} \frac{z^q}{x^q} = \frac{1}{i!} \left(1+\frac{z}{x}\right)^i = \frac{1}{i!} \left(\frac{x+z}{x}\right)^i.$$ This gives for the generating function of the original sum by cancellation the value $$n! x^i (1-x)^{n-i} \frac{1}{(n-i)!} \frac{1}{i!} \left(\frac{x+z}{x}\right)^i \left(\frac{1-x-z}{1-x}\right)^{n-i} = {n\choose i} (x+z)^i (1-x-z)^{n-i}.$$ The conclusion is that the sum is given by $$r! [z^r] {n\choose i} (x+z)^i (1-x-z)^{n-i} = r! [(w-x)^r] {n\choose i} w^i (1-w)^{n-i}.$$ But this is precisely $(B^n_i)^{(r)}(x)$ by the definition of the Taylor series.

Marko Riedel
  • 61,317