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  • one year ago

Assume that \( 1a_1+2a_2+\cdots+na_n=1, \) where the \(a_j\) are real numbers. As a function of \(n\), what is the minimum value of \(1a_1^2+2a_2^2+\cdots+na_n^2?\) I tried 1/5 and then 5, but none of them work. Please help! Thanks!

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  1. anonymous
    • one year ago
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    I'm pretty sure you can use Lagrange multipliers to solve this. Let \(S_1=\sum\limits_{j=1}^nja_j\) and \(S_2=\sum\limits_{j=1}^nj{a_j}^2\). You're looking to minimize \(S_2\) with respect to the constraint \(S_1\). The Lagrangian is given by \[\mathcal{L}(a_1,\ldots,a_n,\lambda)=S_2+\lambda S_1\] Take the gradient and set equal to \(0\): \[\left.\begin{align*} \frac{\partial}{\partial a_1}\left[S_2+\lambda S_1\right]&=2a_1+\lambda\\[2ex] \frac{\partial}{\partial a_2}\left[S_2+\lambda S_1\right]&=4a_2+2\lambda\\[2ex] &\vdots\\ \frac{\partial}{\partial a_n}\left[S_2+\lambda S_1\right]&=2na_n+n\lambda\\[2ex] \frac{\partial}{\partial \lambda}\left[S_2+\lambda S_1\right]&=S_1 \end{align*}\right\}=0\quad(\text{each})\] The first \(n\) equations tell you that \(a_j=-\dfrac{\lambda}{2}\). Plugging this into the original constraint, you get \[S_1=\sum_{j=1}^n ja_j=-\frac{\lambda}{2}\sum_{j=1}^n j=-\frac{\lambda n(n+1)}{4}=1\]or \(\lambda=-\dfrac{4}{n(n+1)}\). This means \(S_2\) is minimized when \[a_j=-\frac{\lambda}{2}=\frac{2}{n(n+1)}\] Compute the sum: \[\begin{align*}\sum_{j=1}^n j{a_j}^2&=\sum_{j=1}^nj\left(\frac{2}{n(n+1)}\right)^2\\[1ex] &=\frac{4}{n^2(n+1)^2}\sum_{j=1}^nj\\[1ex] &=\frac{4}{n^2(n+1)^2}\cdot\frac{n(n+1)}{2}\\[1ex] &=\frac{2}{n(n+1)}\end{align*}\]

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