In theoretical analysis and also numerical work, we often would like to approximate complicated functions with ones that are more easily managed. It turns out that, given a continuous function $f: [0,1] \rightarrow \mathbb{R}$, the sequence of Bernstein polynomials given by $p_n(x)=\sum_{k=0}^n {n \choose k}f(\frac{k}{n})x^k (1-x)^{n-k}$ converges uniformly to $f$ as $n \rightarrow \infty$. Here ${n \choose k}$ is the familiar binomial coefficient, given by ${n \choose k} = \frac{n!}{k!(n-k)!}$
Before validating this assertion with a proof, it is worthwhile to explain what might have motivated it in the first place. Indeed, the quantity ${n \choose k} x^k (1-x)^{n-k}$ is simply the probability of $k$ successes in $n$ Bernoulli trials, where a success occurs with probability $x$.
We can thus view ${n \choose k} x^k (1-x)^{n-k}$ as the probability of getting $k$ heads in $n$ flips of a coin (not necessarily fair), where the probability of a head is $x$. And $f(\frac{k}{n})$ can be seen as representing the payout when exactly $k$ heads are flipped. The expected value of the payout is thus $p_n(x)=\sum_{k=0}^n {n \choose k}f(\frac{k}{n})x^k (1-x)^{n-k}$.
Now, as $n$ gets bigger and bigger, the law of large numbers states that the probability of getting $k$ heads will be very small for most $k$, but large for $k/n$ close to $x$. Thus as $n$ increases, the payout clusters around $f(x)$. So for large $n$, we would expect $p_n(x)$, the expected value of the payout, to be $f(x)$.
Our intuition seems reasonable, but what we have written is hardly a rigorous proof. Rather than providing one, we refer the reader to the standard treatment given in Elementary Classical Analysis by Marsden and Hoffman.
Before validating this assertion with a proof, it is worthwhile to explain what might have motivated it in the first place. Indeed, the quantity ${n \choose k} x^k (1-x)^{n-k}$ is simply the probability of $k$ successes in $n$ Bernoulli trials, where a success occurs with probability $x$.
We can thus view ${n \choose k} x^k (1-x)^{n-k}$ as the probability of getting $k$ heads in $n$ flips of a coin (not necessarily fair), where the probability of a head is $x$. And $f(\frac{k}{n})$ can be seen as representing the payout when exactly $k$ heads are flipped. The expected value of the payout is thus $p_n(x)=\sum_{k=0}^n {n \choose k}f(\frac{k}{n})x^k (1-x)^{n-k}$.
Now, as $n$ gets bigger and bigger, the law of large numbers states that the probability of getting $k$ heads will be very small for most $k$, but large for $k/n$ close to $x$. Thus as $n$ increases, the payout clusters around $f(x)$. So for large $n$, we would expect $p_n(x)$, the expected value of the payout, to be $f(x)$.
Our intuition seems reasonable, but what we have written is hardly a rigorous proof. Rather than providing one, we refer the reader to the standard treatment given in Elementary Classical Analysis by Marsden and Hoffman.
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