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I want to proof that the bernstein-coefficients for $p(x)=x$ on $[a,b]$ are described by $$b_i=a+i\frac{b-a}{n},\ i=0,...,n$$

Where the Bernstein polynomials on $[a,b]$ are defined by $$B_i^n(x;a,b)=(b-a)^{-n}{n\choose i}(b-x)^{n-i}(x-a)^i$$

I think I have to use the partition of unity property, which is $$1=\sum_{i=0}^n B_i^n(x;a,b) ,\ x\in\mathbb{R}$$

My first step is then to write it down: $$p(x)=\sum_{i=0}^n b_i B_i^n(x;a,b)=\sum_{i=0}^n(a+i\frac{b-a}{n})(b-a)^{-n}\binom{n}{i}(b-x)^{n-i}(x-a)^i=$$

[... magic ...]

$$ = x\sum_{i=0}^{n-1} B_i^{n-1}(x;a,b) = x$$

The "magic" part probably needs some 1-1 trick in an exponent or something like that, but I can't see how. Here is a proof of the same thing but on [0,1] which seems straightforward enough but I can't bring my formula into the same shape.

Mike Pierce
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rtur
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1 Answers1

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I've figured it out, in the end it was too simple to see ...

Here is the proof:

$$p(x)=\sum_{i=0}^n[(a+i\frac{b-a}{n})B_i^n(x;a,b)]=$$ $$=a\sum_{i=0}^nB_i^n(x;a,b)+\sum_{i=0}^n[(b-a)\frac{i}{n}B_i^n(x;a,b)]=$$ $$=a+\sum_{i=1}^n[(b-a)^{-(n-1)} \binom{n-1}{i-1}(b-x)^{n-1-(i-1)}(x-a)^{i-1}](x-a)=$$ $$=a+(x-a)\sum_{i=0}^{n-1}[B_{i}^{n-1}(x;a,b)]=x$$

rtur
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