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proof of Weierstrass approximation theorem
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(Proof)
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To simplify the notation, assume that the function is defined on the interval $[0,1]$ . This involves no loss of generality because if $f$ is defined on some other interval, one can make a linear change of variable which maps the domain of $f$ to $[0,1]$ .
Let us start by demonstrating a few special cases of the theorem, starting with the case $f(x) = 1 - \sqrt{1 - x}$ . In this case, we can use the ancient Babylonian method of computing square roots to construct polynomial approximations. Define the polynomials $P_0, P_1, P_2, \ldots$ recursively as $$P_0 (x) = 0$$ $$P_{n+1} (x) = {1 \over 2} \left( P_n(x)^2 + x \right)$$ It is an obvious
consequence of this definition that, if $0 \le x \le 1$ then $0 \le P_n (x) \le 1$ for all $n$ . It is equally obvious that each $P_n$ is a monotonically increasing function on the interval $[0,1]$ . By subtracting the recursion from itself, cancelling, and factoring, we obtain the relation $$P_{n+2} (x) - P_{n+1} (x) = {1 \over 2} (P_{n+1} (x) + P_n (x)) (P_{n+1} (x) - P_n (x))$$ From this relation, we conclude that $P_{n+1} (x) \ge P_n (x)$ for all $n$ and all $x$ in $[0,1]$ . This implies that $\lim_{n \to \infty} P_n (x)$ exists for all $x$ in $[0,1]$ . Taking the limit of both sides of the recursion that defines $P_n$ and simplifying, one sees that $\lim_{n \to \infty} P_n (x) = 1 - \sqrt{1-x}$ . The relation also implies that $P_{n+1} (x) - P_n (x)$ is also a monotonically increasing function of $x$ in the interval $[0,1]$ for all $n$ . Therefore, $$P_{n+1} (x) - P_n (x) \le P_{n+1} (1) - P_n (1)$$ Summing over $n$ and cancelling, one sees that $$P_{m} (x) - P_n (x) \le P_{m} (1) - P_n (1)$$ whenever $m > n$ . Taking the limit as $m$ approaches infinity, one concludes that $$1 - \sqrt{1-x} - P_n(x) \le 1 - P_n (1)$$ Since the $P_n$ 's converge, for any $\epsilon > 0$ , there exists an $n$ such that $1 - P_n (1) < \epsilon$ . For this value of $n$ ,
$|f(x) - P_n (x)| < \epsilon$ , so the Weierstrass approximation theorem holds in this case.
Next consider the special case $f(x) = |x-c|$ , where $0 \le c \le 1$ . A little algebra shows that $$\sqrt{(x-c)^2 + \epsilon^2 / 4} - |x-c| \le \epsilon / 2$$ By the case of the approximation theorem already proven, there exists a polynomial $P$ such that $$|\sqrt{(x-c)^2 + \epsilon^2 / 4} - P(x)| < \epsilon / 2$$ when $x \in [0,1]$ . Combining the last two inequalities and applying the triangle
inequality, one sees that $|f(x) - P(x)| < \epsilon$ , so the Weierstrass approximation theorem holds in the case $f(x) = |x-c|$ .
A corollary of the result just proven is the Weierstrass appriximation theorem for piecewise linear functions. Any piecewise linear function $\phi$ can be expressed as $$\phi(x) = b + \sum_{m=0}^N a_m |x-c_m|$$ for suitable constants $a_0, \ldots , a_N, b, c_0, \ldots , c_N$ . By the result just proven, there exist polynomials $P_0, \ldots,P_N$ such that $$\left| a_m |x - c_m| - P_m \right| < \epsilon / N$$ By the triangle inequality, $$\left| \phi (x) - b - \sum_{m=0}^N P_m \right| < \epsilon$$
Having succeeded in proving all these special cases, we now have the courage to attack the general theorem. In light of the case just proven, it suffices to show that if $f$ is continuous on [0,1] then for all $\epsilon > 0$ there exists a piecewise-linear function $\phi$ such that for all $x$ in $[0,1]$ , $|f(x) - \phi (x)| < \epsilon / 2$ . For, if such a function exists, then there also exists a polynomial $P$ such that $|\phi(x) - P(x)| < \epsilon / 2$ , but then $|f(x) - P(x)| < \epsilon$ .
Since the interval $[0,1]$ is compact, $f$ is uniformly continuous. Hence, for all $\epsilon > 0$ there exists an integer $N$ such that $|f(x) - f(y)| < \epsilon / 2$ whenever $|x - y| \le 1/N$ .
Define $\phi$ by the following two conditions: If $m$ is an integer between $0$ and $N$ , $\phi (m/N) = f (m/N)$ . On any interval $[m/N,(m+1)/N]$ , $\phi$ is linear.
For every point $x$ in the interval $[0,1]$ , there exists an integer $m$ such that $x$ lies in the subinterval $[m/N,(m+1)/M$ . Since a linear function is bounded by its values at the endpoints, $\phi(x)$ lies between $\phi(m/N) = f(m/N)$ and $\phi((m+1)/N) = f((m+1)/N)$ . Since $|f(m/N) - f((m+1)/N)| \le \epsilon/2$ , it follows that $|f(m/N) - \phi(x)|
\le \epsilon/2$ . Because $|x - m/N| \le 1/N$ , it is also true that $|f(m/N) - f(x)| \le \epsilon/2$ . Hence, by the triangle inequality, $|f(x) - \phi(x)| < \epsilon$ .
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"proof of Weierstrass approximation theorem" is owned by rspuzio. [ full author list (3) ]
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Cross-references: endpoints, bounded, subinterval, point, integer, uniformly continuous, compact, continuous, piecewise, triangle inequality, inequalities, approximation theorem, algebra, Weierstrass approximation theorem, converge, infinity, summing, sides, limit, implies, relation, monotonically increasing, consequence, obvious, approximations, polynomial, Babylonian method of computing square roots, theorem, domain, maps, variable, interval, function
This is version 6 of proof of Weierstrass approximation theorem, born on 2004-09-05, modified 2008-12-22.
Object id is 6145, canonical name is ProofOfWeierstrassApproximationTheorem.
Accessed 12666 times total.
Classification:
| AMS MSC: | 41A10 (Approximations and expansions :: Approximation by polynomials) |
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Pending Errata and Addenda
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