Question

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!

Answer 1

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:
\[\begin{align*}\sum_{j=1}^n j{a_j}^2&=\sum_{j=1}^nj\left(\frac{2}{n(n+1)}\right)^2\\[1ex]
&=\frac{4}{n^2(n+1)^2}\sum_{j=1}^nj\\[1ex]
&=\frac{4}{n^2(n+1)^2}\cdot\frac{n(n+1)}{2}\\[1ex]
&=\frac{2}{n(n+1)}\end{align*}\]