Given a function $f(x) \in C[0,1]$, we define the $n$th Bernstein Operator $T_n$ by

$(T_nf)(t) = \sum_{k=0}^n {{n}\choose{k}}f\left(\frac{k}{n}\right)t^k(1 - t)^{n - k}$.

I'm trying to code this in Mathematica, but I don't see how it can be done to take an unknown function as an argument. My current working code is

b[f[x_]_, n_, t_] := Sum[(n!/(k!*(n - k)!))*f(k/n)*t^k*(1 - t)^(n - k), {k, 0, n}]

but this doesn't seem to work. Can anyone offer a way to take an undefined function as an input?

Edit: Though the code below in the comments works spectacularly, the code also works as-is, so long as one remembers to change $f(k/n)$ to $f[k/n]$, so that Mathematica knows to evaluate $f$ at $k/n$.

  • 2
    $\begingroup$ bop[n_Integer, f_, t_] := Sum[Binomial[n, k] Function[t, f][k/n] t^k (1 - t)^(n - k), {k, 0, n}], and then try bop[5, Exp[t], t]. $\endgroup$ Mar 28, 2018 at 20:01
  • $\begingroup$ That worked perfectly! Thank you! $\endgroup$ Mar 28, 2018 at 20:11
  • 2
    $\begingroup$ Also for reference: BernsteinBasis[] is built-in, so you can modify my code to shorten it further. $\endgroup$ Mar 28, 2018 at 20:15
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    – rhermans
    Jun 16, 2018 at 10:44

1 Answer 1


Using a slightly different notation, consistent with mathworld and literature

BernsteinOperator[n_Integer][f_, x_Symbol] := Sum[
  Function[x, f][i/n] BernsteinBasis[n, i, x]
  , {i, 0, n}

PiecewiseExpand[BernsteinOperator[2][a Exp[b x], x]]

Mathematica graphics

   BernsteinOperator[9][Exp[x], x]
 , {x, -0.5, 1.5}
 , PlotTheme -> "Scientific"
 , PlotStyle -> {Gray, Red}

Mathematica graphics


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