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Problem 1130

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  1. vf_68e7f59e503afe2d

    theoretical

    Erdős Problem #1130: declared status 'proved'. Formalized: no. For x1,,xn[1,1]x_1,\ldots,x_n\in [-1,1] letlk(x)=ik(xxi)ik(xkxi),l_k(x)=\frac{\prod_{i\neq k}(x-x_i)}{\prod_{i\neq k}(x_k-x_i)},which are such that lk(xk)=1l_k(x_k)=1 and lk(xi)=0l_k(x_i)=0 for iki\neq k. Let x0=1x_0=-1 and xn+1=1x_{n+1}=1 andΥ(x1,,xn)=min0inmaxx[xi,xi+1]klk(x).\Upsilon(x_1,\ldots,x_n)=\min_{0\leq i\leq n}\max_{x\in[x_i,x_{i+1}]} \sum_k \lvert l_k(x)\rvert.Is it true thatΥ(x1,,xn)logn?\Upsilon(x_1,\ldots,x_n)\ll \log n?Describe which choice of xix_i maximise Υ(x1,,xn)\Upsilon(x_1,\ldots,x_n). Current best: Erd\H{o}s [Er47] could proveΥ(x1,,xn)<n.\Upsilon(x_1,\ldots,x_n)< \sqrt{n}.Erd\H{o}s thought that the maximising choice is characterised by the property that the sumsmaxx[xi,xi+1]klk(x)\max_{x\in [x_i,x_{i+1}]}\sum_k \lvert l_k(x)\rvertare all equal for 0in0\leq i\leq n (where x0=1x_0=-1 and xn+1=1x_{n+1}=1), which would be the same (conjectured) characterisation as [1129]. Prize: no. Tags: analysis, polynomials.

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