How do I apply the distributive law to ∑(k xi + j yi)?

[dE_logics]

If I have ∑(k x_i + j y_i)...how will I apply the distributive law on it?...I mean how do you split this notion?

 

[tiny-tim]

Hi dE_logics!

∑(k x_i + j y_i)

= ∑(k x_i) + ∑(j y_i)

= k ∑(x_i) + j ∑(y_i)

(if the ∑ is over infinitely many terms, you may have to be careful about convergence …

but if for example all the terms are positive, then there's no difficulty )

 

[dE_logics]

Hi dE_logics!

∑(k x_i + j y_i)

= ∑(k x_i) + ∑(j y_i)

= k ∑(x_i) + j ∑(y_i)

(if the ∑ is over infinitely many terms, you may have to be careful about convergence …

but if for example all the terms are positive, then there's no difficulty )

So we won't get ∑(k x_i) + ∑(j y_i)...that was too a possibility.

So let's take an e.g. -

x1 = 1, x2 = 2, x3 = 7, x4 = 1
y1 - 19, y2 = 8, y3 = -10, y4 = 0
k = j= 3

∑(k x_i + j y_i) gives -30

k ∑(x_i) + j ∑(y_i) = -30

and

∑(k x_i) + ∑(j y_i) = -30

So both of the solutions are true...does everyone agree?...I mean -

∑(k x_i) + ∑(j y_i) = k ∑(x_i) + j ∑(y_i)

 

[HallsofIvy]

So we won't get ∑(k x_i) + ∑(j y_i)...that was too a possibility.

??Yes, we will, that was Tiny-tim's first step! But you asked about the distributive law and that hasn't been used yet, so then he factored k and j out.

So let's take an e.g. -

x1 = 1, x2 = 2, x3 = 7, x4 = 1
y1 - 19, y2 = 8, y3 = -10, y4 = 0
k = j= 3

∑(k x_i + j y_i) gives -30

k ∑(x_i) + j ∑(y_i) = -30

and

∑(k x_i) + ∑(j y_i) = -30

So both of the solutions are true...does everyone agree?...I mean -

∑(k x_i) + ∑(j y_i) = k ∑(x_i) + j ∑(y_i)

Yes, of course.

 

[dE_logics]

Oh...I didn't see that...thanks.