Given $P(x) = ax^3 + bx^2 + cx + d$ where $ \vert P(x) \vert \le 1\; \forall \vert x \vert \le 1$ Prove: $$ \vert a \vert + \vert b \vert + \vert c \vert + \vert d \vert \le 7$$ Hopes you guys can help me with this problem. Thank you so much!!!!!!
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I'm thinking of using Taylor Series as the graph of the cubic polynomial looks like the sine function graph in [-1,1]. But that just a graph... I really don't know where to start – HAHAHAHHA Oct 27 '20 at 13:36
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You should clearly state in your question that you're talking about real polynomials on the real interval $[-1,1]$, not complex polynomials on the closed unit disk $|x|\le1$. The answer below seems to assume the latter, but then, $|P(x)|\le1$ would be a far stronger assumption, allowing sharper estimates. – Oct 27 '20 at 16:29
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You should look into Chebyshev polynomials. $x^3-\frac{3}{4}x$ can be shown to be that unique monic cubic polynomial that has the least $|p|\infty:=\max{x\in[-1,1]}|p(x)|$. Thus it follows that $4x^3-3x$ has the largest leading coefficient among all cubics with $|p|_\infty=1$. In fact the sum of its coefficients is $7$. – Chrystomath Oct 30 '20 at 17:24