Polynomials
How would one prove the claim in the solution? Why must it be between -n² and n²?
f_1 is first derivative btw.
11 Replies
@Apu
Note for OP
+solved @user1 @user2... to close the thread when your doubt is solved. Mention the users who helped you solve the doubt. This will be added to their stats.Cherbyshev polynomial or something?
I have no clue.
@SirLancelotDuLac pls expand if you can.
I think $n^{2}$ is a broader range. I searched this up and yeah, pretty much this idea also gives the same:
Substitute x=$cos(\theta)$, then the polynomial can be written as $a{0}+a{1}cos(\theta)+a{2}cos(2 \theta)+...$ (Since cos($n \theta$) is a polynomial in $cos(\theta)$ of the degree n). Now $1 \leq a{n}cos(n \theta)+a_{n-1}cos((n-1) \theta)+... \leq 1$ and maximum of cosine is 1.
Deducing from here this gives that the derivative lies between -(n^2+n)/2 and (n^2+n)/2.
SirLancelotDuLac
How are the coefficients going to be the same?
a_i here is different from a_i in the question.
Oh wait yeah this would not be correct.
I did look up Chebyshev polynomials, but I couldn't figure out how to apply them.
The only thing I could think of was the cos substitution. Idk where to go from here.
Ok, I'll try it out.