How to increase the correlation? I have three vectors of numbers with the same dimensionality, $A$,$B$ and $C$. What is the most suitable number $x$, which maximizes the correlation of $A$ and $B+xC$ . To what extend can I increase the correlation. 
Thanks 
 A: (I interpret correlation as described here.) Since the correlation does not change if you displace any of these vectors by a multiple of the constant vector $\mathbf{1}$ we can assume that $$\langle A, \mathbf{1} \rangle = \langle B, \mathbf{1} \rangle = \langle C, \mathbf{1} \rangle = 0.$$  (That is, we can assume that each of these vectors has zero mean.)  Then the correlation between $A$ and $B+xC$ is 
$$
\operatorname{Corr}(A, B+xC) = \frac{\langle A, B + xC \rangle}{||A|| \cdot ||B + x C||}
$$
The derivative in $x$ of this expression is
$$
\frac{\langle A,C \rangle \langle B, B \rangle - \langle A, B \rangle \langle B, C \rangle + x (\langle A, C \rangle \langle B, C \rangle - \langle A, B \rangle \langle C, C \rangle)}{||A|| \cdot ||B + x C||^3}
$$
and therefore if the correlation is maximal or minimal then
$$
x = \frac{\langle A,C \rangle \langle B, B \rangle - \langle A, B \rangle \langle B, C \rangle}{\langle A, B \rangle \langle C, C \rangle - \langle A, C \rangle \langle B, C \rangle} = \frac{||B||}{||C||} \cdot \frac{\operatorname{Corr}(A,C) - \operatorname{Corr}(A,B) \operatorname{Corr}(B,C)}{\operatorname{Corr}(A,B) - \operatorname{Corr}(A,C) \operatorname{Corr}(B,C)}.
$$
If the correlation has a maximum then the correlation as function in $x$ never reaches a minimum and vice versa.  Both situations can occur (e.g. replace $A$ by $-A$).
Edit: (Exposition of the derivative.)  Substitute $x \leftarrow x+t$ in the expression for the correlation to obtain:
$$
\frac{\langle A, B + xC \rangle + t \langle A, C \rangle}{||A|| \cdot \left(||B + x C||^2 + 2t \langle B + xC, C \rangle + O(t^2)\right)^{1/2}} =
$$
$$
\frac{\langle A, B + xC \rangle + t \langle A, C \rangle}{||A|| \cdot ||B + x C|| \cdot \left(1 + t \frac{\langle B + xC, C \rangle}{||B+xC||^2} + O(t^2)\right)} =
$$
$$
\frac{\langle A, B + xC \rangle + t \langle A, C \rangle}{||A|| \cdot ||B + x C||} \left(1 - t \frac{\langle B + xC, C \rangle}{||B+xC||^2} + O(t^2)\right) =
$$
$$
\frac{\langle A, B + xC \rangle}{||A|| \cdot ||B + x C||} + t \left( \frac{\langle A, C \rangle}{||A|| \cdot ||B + x C||} - \frac{\langle A, B + xC \rangle \langle B + xC, C \rangle}{||A|| \cdot ||B + x C||^3}\right) + O(t^2)
$$
Therefore the derivative in $x$ of the correlation is
$$
\frac{\langle A, C \rangle}{||A|| \cdot ||B + x C||} - \frac{\langle A, B + xC \rangle \langle B + xC, C \rangle}{||A|| \cdot ||B + x C||^3}
$$
which after some rewriting is the expression above.  Note that in particular the $x^2$ term in the numerator cancels.
