Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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.

share|cite|improve this question
up vote 7 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!

share|cite|improve this answer
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
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 $[-1,1]$.

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

Actually you can use Bernstein polynomial formula to approximate any continous functions on [a,b].

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.