Summation Question

[liars_paradox]

Hi, simple question I have here. I was wondering how you would go from the following expression:

 

 

to this one:

 

?

 

 

 

 

[Shadow]

Assuming that n, m are constants:

 

[math]\sum_{j=i+1}^n \frac{1}{m} = (n- (i+1)+1) \cdot \frac{1}{m} = \frac{n-i}{m}[/math]

 

[math]\frac{1}{n}\sum_{i=1}^{n}(1 + \sum_{j=i+1}^n \frac{1}{m}) =\frac{1}{n} \sum_{i=1}^{n}(1 + \frac{n-i}{m}) = [/math]

 

[math] = \frac{1}{n} \cdot n \cdot (1 + \frac{n-i}{m}) = 1 + \frac{1}{n} \frac{n(n-i)}{m} = 1 + \frac{1}{nm} \cdot n(n-i) = [/math]

 

[math] = 1 + \frac{1}{mn} \sum_{i = 1}^{n} (n-i)[/math]

 

Don't know why you would want to simplify only partially though; the above result can still be simplified to [math]\frac{m+n-i}{m}[/math]. Basically all of the simplification made use of the fact that [math]\sum_{i=j}^{k} c = (j-k+1)\cdot c[/math] where [math] c \in \mathbb{R}[/math].

[liars_paradox]

Assuming that n, m are constants:

 

[math]\sum_{j=i+1}^n \frac{1}{m} = (n- (i+1)+1) \cdot \frac{1}{m} = \frac{n-i}{m}[/math]

What happened to the 2 here? Did you just simply drop it or did something else happen to it?

 

[math]\frac{1}{n}\sum_{i=1}^{n}(1 + \sum_{j=i+1}^n \frac{1}{m}) =\frac{1}{n} \sum_{i=1}^{n}(1 + \frac{n-i}{m}) = [/math]

 

[math] = \frac{1}{n} \cdot n \cdot (1 + \frac{n-i}{m}) = 1 + \frac{1}{n} \frac{n(n-i)}{m} = 1 + \frac{1}{nm} \cdot n(n-i) = [/math]

 

[math] = 1 + \frac{1}{mn} \sum_{i = 1}^{n} (n-i)[/math]

 

Don't know why you would want to simplify only partially though; the above result can still be simplified to [math]\frac{m+n-i}{m}[/math]. Basically all of the simplification made use of the fact that [math]\sum_{i=j}^{k} c = (j-k+1)\cdot c[/math] where [math] c \in \mathbb{R}[/math].

 

And, what about your formula: [math]\sum_{i=j}^{k} c = (j-k+1)\cdot c[/math]? Wouldn't it be more like [math]\sum_{i=j}^{k} c = (k-j+1)\cdot c[/math], where j is subtracted from k as opposed to it being the other way around?

Edited January 25, 2011 by liars_paradox

[Shadow]

I can't see a two anywhere. If you mean the two ones, don't forget that the minus sign will flip the signs inside the parenthesis, so [math](n- (i+1)+1)[/math] will become [math](n - i - 1+1)[/math].

 

And yes, my mistake; it should be [math] \sum_{i=j}^{k} c = (k-j+1)\cdot c [/math].