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I'm going through some problems and I'm really stumped on this one. The questions says that

Given $f(x)=|x|$, show that there is a sequence of (real) polynomials $P_n(x)$ with $P_n(0)=0$ that converge uniformly to $f(x)$ on the interval $[-1,1]$.

I think an application of the Weierstrass theorem is in order, but I don't know how to apply it here and so I'll need some help.
Thanks.

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2 Answers 2

up vote 6 down vote accepted

As you correctly guessed, you can use that by Weierstrass's theorem there is a sequence of polynomials $Q_n(x)$ uniformly converging to $f$ on $[-1,+1]$.
Done? Not quite: those $Q_n$'s might not satisfy $Q_n(0)=0$.
Well, then give them a little push that will force them to comply: do you see what you have to add to each of them to obtain $P_n$ and why the push is little (and becomes littler and littler) ?
And do you see why the new sequence of $P_n$'s will still converge to $f$ uniformly because of the mentioned littleness?
And do you see that the exact formula for $f$ is a red herring and that only the fact that $f(0)=0$ is relevant?

Yes, I'm sure you'll see all that after a short moment a reflexion. Good luck!

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Thanks for your response. If I define $P_n(x) = Q_n(x)$, then one property is satisfied. To get $P_n(0)=0$, I need $Q_n(0)=0$. But if $Q_n(x)\rightarrow |x|$, then $Q_n(0)\rightarrow 0?$ If so then I can define $P_n(x)=Q_n(x)-Q_n(0)$. Is this right? –  Cindy Nov 29 '11 at 9:30
1  
Dear @Cindy, your definition of $P_n(x)$ is the right one: congratulations! To finish the proof, you have to show uniform convergence to $0$ of $|P_n-f|=|Q_n-Q_n(0)-f|=|(Q_n-f)-Q_n(0)|$ and for that you can use the triangle inequality. –  Georges Elencwajg Nov 29 '11 at 13:20
    
Thanks very much. –  Cindy Nov 29 '11 at 15:55

As some proofs of the Weierstrass Approximation Theorem use this result for its proof, I am somewhat unsatisifed with using the Weierstrass Approximation Theorem to prove it.

The Taylor series for $\sqrt{1-x}$ is $$ \sqrt{1-x }=1-{1\over 2}x-{1\over 2\cdot4}x^2-{1\cdot3\over 2\cdot4\cdot6}x^3-\cdots. $$ This converges uniformly for $0\le x\le 1$.

From this, we may represent $|x|$ with (replace "$x$" with "$1-x^2$")

$$ |x|=1-{1\over 2}(1-x^2)-{1\over 2\cdot4}(1-x^2)^2-{1\cdot3\over 2\cdot4\cdot6}(1-x^2)^3-\cdots. $$ This will converge uniformly for $0\le1-x^2\le 1$; that is, for $-1\le x\le 1$.

You can then modify things to produce polynomials $P_n$ with $P_n(0)=0$ that converge uniformly to $|x| $ on $[-1,1]$.

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It is nice that you avoid Weierstrass's theorem for fear of a vicious circle: +1. –  Georges Elencwajg Nov 30 '11 at 7:08

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