# Challenge Question

#### Superleo

This question is hard. Beginner can walk away:ninja:
I doubt anyone can solve this

Show that if |x-a|<1 and p(x)=c0+c1x+...+cnx^n is a polynomial of degree n then |p(x)|<=C(n+1)(|a|+1)^n where C=max({|c0|,|c1|,..., |cn|})

#### v8archie

Math Team
Hint: $|p(x)| \le C|1+x+\ldots+x^n| = C\left|{1 - x^{n+1} \over 1-x}\right|$

• 1 person

#### skipjack

Forum Staff
Hi Superleo! Do you think you could use latex (math tags)?

#### zylo

$$\displaystyle |P(x)|\leq C|1+|x|+|x|^{2}+..x^{n}|=\frac{1-|x|^{n+1}}{1-|x|}\\ ||x|-|a||\leq |x-a|<1\\ |a|-1\leq |x|\leq |a|+1\\$$
$$\displaystyle \frac{1}{1-x}\leq- \frac{1}{|a|}\\$$
$$\displaystyle \frac{1-|x|^{n+1}}{1-x}\leq \frac{1}{|a|}(-1+|x|^{n+1})\\\\$$
$$\displaystyle |x|^{n+1}\leq (1+|a|)^{n+1}=1+(n+1)|a|+\frac{1}{2}(n+1)n|a|^{2}+...\\\\ \leq 1+(n+1)|a|(1+|a|)^{n}\\\\$$
$$\displaystyle \frac{1-|x|^{n+1}}{1-x}\leq (n+1)(1+|a|)^{n}$$

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• 1 person

#### zylo

$$\displaystyle |P(x)|\leq C|1+|x|+|x|^{2}+..x^{n}|=C\frac{1-|x|^{n+1}}{1-|x|}\\ ||x|-|a||\leq |x-a|<1\\ |a|-1\leq |x|\leq |a|+1\\$$
$$\displaystyle \frac{1}{1-|x|}\leq- \frac{1}{|a|}\\$$
$$\displaystyle \frac{1-|x|^{n+1}}{1-|x|}\leq \frac{1}{|a|}(-1+|x|^{n+1})\\\\$$
$$\displaystyle |x|^{n+1}\leq (1+|a|)^{n+1}=1+(n+1)|a|+\frac{1}{2}(n+1)n|a|^{2}+...\\\\ \leq 1+(n+1)|a|(1+|a|)^{n}\\\\$$
$$\displaystyle \frac{1-|x|^{n+1}}{1-|x|}\leq (n+1)(1+|a|)^{n}$$
Came back to see who actually thanked me and caught two typo's, one trivial and one that should be corrected.
The trivial one is a missing C.
In a few places 1/(1-x) should be 1/(1-|x|)

The corrections are made in the above quote.
Thanks Ould Youbba. Without your thanks I would have missed the typo's

I originally did the whole thing in plain text and then resubmitted it in Latex to satisfy skipjack. Subsequently skipjack deleted the plain text version.

#### al-mahed

I originally did the whole thing in plain text and then resubmitted it in Latex to satisfy skipjack. Subsequently skipjack deleted the plain text version.
This satisfies the rest of the community as well. #### zylo

This satisfies the rest of the community as well. Good