Assume that $1 a_{1} + 2 a_{2} + \dots + n a_{n} = 1,$ $1a_1+2a_2+\cdots+na_n=1,$ where the $a_{j}$ $a_j$ are real numbers. As a function of $n$ $n$ , what is the minimum value of $1 a_{1}^{2} + 2 a_{2}^{2} + \dots + n a_{n}^{2} ?$ $1a_1^2+2a_2^2+\cdots+na_n^2?$

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Answer 1 · 2017-05-21T00:31:23

$\frac{2}{n (n + 1)}$ $2/(n(n+1)$

$n \sum k = 1 k a_{k} = 1$ $sum_(k=1)^n k a_k = 1$ is the restriction and $n \sum k = 1 k a_{k}^{2}$ $sum_(k=1)^n k a_k^2$ is the objective function

The Lagrangian reads

$L (a, λ) = n \sum k = 1 k a_{k}^{2} + λ (n \sum k = 1 k a_{k} - 1)$ $L(a,lambda)=sum_(k=1)^n k a_k^2+lambda(sum_(k=1)^n k a_k - 1)$

The stationary conditions

$\frac{\partial L}{\partial a_{k}} = 2 k a_{k} + λ k = 0$ $(partialL)/(partial a_k) = 2ka_k+lambda k = 0$ for $k = 1, 2, \dots, n$ $k=1,2,cdots,n$

and

$\frac{\partial L}{\partial λ} = n \sum k = 1 k a_{k} - 1 = 0$ $(partialL)/(partial lambda) = sum_(k=1)^n k a_k- 1 = 0$

so

$a_{k} = - \frac{λ}{2}$ $a_k = -lambda/2$ and

$n \sum k = 1 k (- \frac{λ}{2}) = 1$ $sum_(k=1)^n k ( -lambda/2)= 1$ or

$λ = - \frac{4}{n (n + 1)}$ $lambda = -4/(n(n+1))$ so

$a_{k} = \frac{2}{n (n + 1)}$ $a_k = 2/(n(n+1))$

so the minimum value is given by

$n \sum k = 1 k (\frac{4}{{(n (n + 1))}^{2}}) = \frac{2}{n (n + 1)}$ $sum_(k=1)^n k (4/(n(n+1))^2) = 2/(n(n+1))$

Assume that 1a_1+2a_2+\cdots+na_n=1, 1a1+2a2+⋯+nan=1, 1a_1+2a_2+\cdots+na_n=1, where the a_jaja_j are real numbers. As a function of nnn, what is the minimum value of 1a_1^2+2a_2^2+\cdots+na_n^2?1a21+2a22+⋯+na2n?1a_1^2+2a_2^2+\cdots+na_n^2?