I have
$$\sum_k^n P_k x_k$$
Am I allowed to split it up into two sums so I have it like
$$\sum_k^n P_k \sum_k^nx_k$$
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2$\begingroup$ No. But since $P_{n}$ doesn't depend on the sum (of $k$), you can take it out, and it becomes $P_{n}\sum_{k=1}^{n}x_{k}$. $\endgroup$ – Vinícius Novelli Oct 5 '14 at 3:41
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1$\begingroup$ In $\sum_k^nP_nx_k$, $P_n$ doesn't depend on $k$ so can be factored out. $\endgroup$ – Kim Jong Un Oct 5 '14 at 3:41
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$\begingroup$ Assuming that $P_n$ is a constant with respect to $k$, the above comments are true. $\endgroup$ – Squirtle Oct 5 '14 at 3:48
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3$\begingroup$ $2 = 1\cdot 1 + 1\cdot 1 \neq (1+1)\cdot(1+1) = 4$. $\endgroup$ – Slade Oct 5 '14 at 6:18
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$$x_1\qquad\qquad\qquad\qquad x_2\qquad\qquad\qquad\qquad x_3\qquad\qquad\qquad\qquad x_4$$
Imagine that the vertical segments are your $P$'s. Then the product of sums is the area of the entire rectangle, whereas the sum of products is only the red area.
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Assuming P depends on k and not n then the summation $$ \sum_k P_k x_k $$ Creates terms like $$ P_1 x_1 + P_2 x_2 + \ldots + P_n x_n $$ But, if you took $P_k$ out then you would have two independent sums: $$ \sum_k P_k \sum_k x_k $$ Producing two separate sums multiplying each other like $$ (P_1 + P_2 + \ldots + P_n) (x_1 + x_2 + \ldots + x_n) $$ And this obviously produces terms like $P_1 x_1, P_1 x_2,\ldots$ or as produced by the double sum written like: $$ \sum_i \sum_j P_i x_j $$ So the original summation is definitely not the same as this double sum.
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http://tutorial.math.lamar.edu/Classes/CalcI/SummationNotation.aspx
^^^That offers a nice summary of summation notation and its properties.
To answer your question specifically, you should not split this into two different summations. Rather, notice that P does not depend on k. You can treat P as a constant and simply move it outside the summation so that it multiples the result of your summation of $x_k$, which does depend on your variable k.
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$\begingroup$ actually I made a typo P depends on k and n. I ask because I know that the the sum with the $P_nk is always equal to 1 for all n and k. But thanks for the response, it does make sense with what I wrote! $\endgroup$ – Jeff Oct 5 '14 at 4:09
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$\begingroup$ That link answers my question though, I appreciate it thanks $\endgroup$ – Jeff Oct 5 '14 at 4:10
