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Consider the problem of finding the values of $\alpha_1, \alpha_2, ..., \alpha_k$, subject to constraints, such that the following equation is satisfied

\begin{equation} \alpha_1 x_1 + \alpha_2 x_2 + \dots \alpha_k x_k = P \end{equation}

where all $x_i$'s and $P$ are real numbers.

For example, the constraint of uniform scaling of the $x_i$'s would be setting $\alpha_i = \frac{P}{\sum_i x_i}$.

Now I wish to scale the $x_i$'s such that the ratio between $\alpha$'s is equal to the ratio between the respective $x$'s, that is:

\begin{equation} \frac{\alpha_m}{\alpha_n} = \frac{x_m}{x_n} \end{equation} for all $m, n$.

A solution is to set:

\begin{equation} \alpha_i = \frac{x_i P}{x_1^2 + x_2^2 + \dots + x_k^2} \end{equation}

What is this called? Is there anyway to find this solution quickly without knowing it before hand?

Can it be generalised?

Xander Henderson
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Alex
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1 Answers1

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Do you mean something like this? \begin{eqnarray*} \frac{1}{\alpha _{j}}P &=&\frac{1}{\alpha _{j}}\{\alpha _{1}x_{1}+\cdots +\alpha _{k}x_{k}\}=\frac{\alpha _{1}}{\alpha _{j}}x_{1}+\frac{\alpha _{2}}{% \alpha _{j}}x_{2}+\cdots x_{j}+\cdots \frac{\alpha _{k}}{\alpha _{j}}x_{k} \\ &=&\frac{x_{1}^{2}}{x_{j}}+\frac{x_{2}^{2}}{x_{j}}+\cdots \frac{x_{k}^{2}}{% x_{j}}=\frac{1}{x_{j}}\sum_{j=1}^{k}x_{j}^{2}\Rightarrow \alpha _{j}=\frac{% x_{j}}{\sum_{j=1}^{k}x_{j}^{2}}P \end{eqnarray*} In the same way \begin{equation*} x_{j}=\frac{\alpha _{j}}{\sum_{j=1}^{k}\alpha _{j}^{2}}P \end{equation*} You can look upon $\{\alpha _{1},\cdots ,\alpha _{k}\}$ and $\{x_{1},\cdots ,x_{k}\}$ as $k$-dimensional vectors, so \begin{equation*} \mathbf{\alpha \cdot x}=P \end{equation*} Note that the component of $\mathbf{x}$ orthogonal to $\mathbf{\alpha }$ is arbitrary and vice versa. With your additional conditions you find that $% \mathbf{\alpha }$ is directed along $\mathbf{x}$ and for their norms \begin{equation*} |\mathbf{x}||\mathbf{\alpha }|=P \end{equation*}

Urgje
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