7
$\begingroup$

If I have two expressions with sums in them, like this:

$$\begin{align*} b&=\frac{\sum_{i} (x_i - \bar{x})(y_i -\bar{y})}{\sum_{i}(x_i -\bar{x})^2}\\ r&=\frac{\sum_{i} (x_i - \bar{x})(y_i -\bar{y})}{\sqrt{\sum_{i}(x_i -\bar{x})^2\sum_{i}(y_i -\bar{y})^2}} \end{align*}$$

and I wanted to produce a simplified expression of $\frac{b}{r}$, how would I do it? The problem is the sums. It seems Mathematica doesn't like the unspecified, unevaluated sums.

Edit: I was expecting to end up with something like this:

$$\sqrt{\frac{\sum_{i}(y_i -\bar{y})^2}{\sum_{i}(x_i -\bar{x})^2}}$$

I've been playing around with Expand, Simplify, FullSimplify. There may just be a way to apply it that I'm missing.

$\endgroup$
12
  • $\begingroup$ Please post your Mma code for the sums $\endgroup$ Aug 9 '12 at 12:28
  • 3
    $\begingroup$ @J.M. That was just in case you were a chemist :D $\endgroup$ Aug 9 '12 at 12:42
  • 3
    $\begingroup$ @Verde, I am a chemist, you insensitive clod! ;P $\endgroup$ Aug 9 '12 at 12:44
  • 4
    $\begingroup$ @Verde I have a file on you too. $\endgroup$ Aug 9 '12 at 18:59
  • 1
    $\begingroup$ @verde There are lots of sausages in there. $\endgroup$ Aug 9 '12 at 19:12
8
$\begingroup$

The absence of a terminal $n$ in the sums suggests you are looking for a combination of symbolic reduction and typography. Let's separate the two, then, by using symbols for the sums, performing the reduction, and then replacing the symbols by whatever we like:

MMA screen shot

$\endgroup$
2
  • 1
    $\begingroup$ Or, one can use PowerExpand[]... $\endgroup$ Aug 9 '12 at 15:47
  • $\begingroup$ @J.M. Interestingly (but understandably), using PowerExpand without explicit assumptions yields a different answer: the square roots are no longer distributed over the quotient. $\endgroup$
    – whuber
    Aug 9 '12 at 15:56
5
$\begingroup$

Here is a completely symbolic manipulation:

ClearAll[x, y, xMean, yMean, n];

varianceX = Sum[(x[i] - xMean)^2, {i, n}]

$\sum _i^n (x(i)-\text{xMean})^2$

varianceY = Sum[(y[i] - yMean)^2, {i, n}]

$\sum _i^n (y(i)-\text{yMean})^2$

coVariance = Sum[(x[i] - xMean) (y[i] - yMean), {i, n}]

$\sum _i^n (x(i)-\text{xMean}) (y(i)-\text{yMean})$

b = coVariance/varianceX;

r = coVariance/Sqrt[varianceX varianceY];

b/r /. {Sqrt[a_ b_] :> Sqrt[a] Sqrt[b]}

$\frac{\sqrt{\sum _i^n (y(i)-\text{yMean})^2}}{\sqrt{\sum _i^n(x(i)-\text{xMean})^2}}$

All I had to do in order to help Mathematica get to the simplified result is to state the rule Sqrt[a_ b_] :> Sqrt[a] Sqrt[b].

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.