Understanding Summation and Latex: A Simple Explanation

[quasar_4]

I feel so silly asking this question, but is (the summation is over n1 from 1 to infinity. I have no idea how to type it with the latex)

$$\sum$$(x1^n1)/n1!*(x^(n-n1))/(n-n1)!

= lim$$(_{n1 \rightarrow \infty})$$ (1 + x/n1)^n1 * lim$$(_{n1 \rightarrow \infty})$$ (1 + x/(n-n1))^(n-n1)

= exp(x1)*exp(x2)

?

I know it's simple, but I'm just not sure what the rules are for products inside sums. And also, how the heck do you learn the whole latex thing? I can't for the life of me it out.

 

[offtheleft]

$$\sum$$ $$\frac{(x1^n^1)}{n1!}$$ $$\ast$$ $$\frac{(x^(^n^-^n^1^))}{(n-n1)!}$$

is this correct so far?

 

[HallsofIvy]

I feel so silly asking this question, but is (the summation is over n1 from 1 to infinity. I have no idea how to type it with the latex)

$$\sum$$(x1^n1)/n1!*(x^(n-n1))/(n-n1)!

= lim$$(_{n1 \rightarrow \infty})$$ (1 + x/n1)^n1 * lim$$(_{n1 \rightarrow \infty})$$ (1 + x/(n-n1))^(n-n1)

= exp(x1)*exp(x2)
?

I know it's simple, but I'm just not sure what the rules are for products inside sums. And also, how the heck do you learn the whole latex thing? I can't for the life of me it out.

Please start by telling us what the problem really is! In your first sum you have "x1" and "x". In second line, only "x" and then in the third line "x1" and "x2".

 

[quasar_4]

oh yes, it does seem that I missed some details (it was late at night :rofl:).

The summation is over n1, from 0 to infinity, and there is an x1 and x2 (they're just constants). So it's just what offtheleft has written, except I forgot to put x2 instead of x on the second term At first I thought that this would just turn into two exponentials, because each of the pieces resemble exponentials, but I'm not so sure now. I don't know much about how sums fit into algebra (having only seen them for the most part in calculus and linear algebra, but never for the sake of learning their properties).

Is it ever the general case that the sum (x*y) = sum(x)*sum(y)? I guess I can answer my own question with a counterexample...

so in that case, how do I simplify my sum? In the end I think the correct answer is supposed to be an (x1+x2).