Summation with index greater than one?

[Craiger1987]

Hi, another question for you fine folks. Just getting into the use of sigma and that thing is super handy. And I know if I wanted to solve for say 1 through 10 I can just use i(i+1)/2. But what if the index number is greater than one? The equation doesn't work anymore. Is there a modification or different equasion to use? Thanks

[studiot]

I know you mention sigma, but the context suggests you might be thinking of mathematical iteration, computer (programming) iteration, linear combinations even simultaneous equations.

 

Can you provide some more detail?

[Craiger1987]

11 minutes ago, studiot said:

I know you mention sigma, but the context suggests you might be thinking of mathematical iteration, computer (programming) iteration, linear combinations even simultaneous equations.

 

Can you provide some more detail?

Yeah I meant sigma, sorry  so with sigma if the number below I is one and the top 10 it's just saying 1+2+3...+9+10 and this can be solved for with i(i+1)/2 instead of manually going through all the numbers. But let's say the bottom number we're something other than one, like a 4, the equasion no longer works for summing the integers. Is there a different equasion for when the beginning number is greater than one?

[studiot]

33 minutes ago, Craiger1987 said:

Yeah I meant sigma, sorry  so with sigma if the number below I is one and the top 10 it's just saying 1+2+3...+9+10 and this can be solved for with i(i+1)/2 instead of manually going through all the numbers. But let's say the bottom number we're something other than one, like a 4, the equasion no longer works for summing the integers. Is there a different equasion for when the beginning number is greater than one?

I am more mystified than ever.

The sigma (notation) is not an equation it is an instruction to add something(s) up.

That something must be a n expression involving the indexing given by the start and stop points attached to the sigma.

You don't solve this you follow the instructions. You solve an equation.

 

There is something called a recurrence relation which you can solve for the nth term.

[Craiger1987]

That's what I mean, all the stuff I've read explains how to simply sum all the integers between the start and stop point using a simple equasion i(I+1)/2, provided the start point is 1. But what if the start point is greater than one? This equasion no longe works. How do you add all the integers between the start and stop point like you could if the start point was 1. Like if the instructions were add everything from 4 to 68 for example. 

I know I could run it for 1-68 and then subtract the sum of 1-3 to get it but I was wondering if there was a different way to do it that didn't require that additional step. 

[Sensei]

in this particular case, if you have equation (or computer subroutine) returning you sum from i....j elements.

But you have to know sum of k....j elements, where k>i

You can use previous sum i...j and subtract sum of i...k-1.

f.e. sum from 1...10 is 55,

Sum from 1..5 is 15

Sum from 6...10 is equal to sum of 1..10 (55) minus sum of 1..5 (15) = 40

ps. I see you mentioned this method in the post above, when I was in the middle of replying..

 

Edited August 31, 2017 by Sensei

[studiot]

What you are calling an equation is the expression for the 'sum to n terms' of some (perhaps infinite) sequence that can be found in and closed expression for n

For instance the sum to n terms for the fourth power of the positive integers is

[math]\sum\limits_1^n {{r^4}}  = n\left( {n - 1} \right)\left( {2n + 1} \right)\left( {3{n^2} + 3n - 1} \right)/30[/math]

In this case yes you must generally do it in stages as you suggest since there must be a first term an a further (n-1) terms to make the formula true.

Although it is written with an equals sign it is really a formula for calculating the sum rather than an equation to be solved for something.

But the sigma may also be used for other purposes, without starting at 0 or 1.

 

For instance

[math]\sum\limits_{i = 10}^{i = 20} {{a_i}x_i^2} \quad means\quad {a_{10}}x_{10}^2 + {a_{11}}x_{11}^2 + {a_{12}}x_{12}^2 + ....... + {a_{15}}x_{15}^2 + .......{a_{20}}x_{20}^2[/math]

 

Again this is a formula for calculating something

Here it means take the square of the 10th variable x₁₀ and multiply it by the 10th coefficient a₁₀ then do this for the 11th variable and coefficient all the way up to the 20th variable and coefficient.

You might do this when calculating a line in a large sparse matrix.

[DrKrettin]

9 hours ago, studiot said:

 

But the sigma may also be used for other purposes, without starting at 0 or 1.

 

The sigma just means Sum. Note that the integral sign is a distorted S (or sigma) to indicate the integral being an infinite series of infinitesimal components. I don't think this is relevant, just interesting. 

[studiot]

1 hour ago, DrKrettin said:

The sigma just means Sum. Note that the integral sign is a distorted S (or sigma) to indicate the integral being an infinite series of infinitesimal components. I don't think this is relevant, just interesting. 

That's what I said, but sum on its own is insufficient. You need information about what to sum, and the full notation contains this.

12 hours ago, studiot said:

The sigma (notation) is not an equation it is an instruction to add something(s) up.

for example the following tells us to  form n increments alpha to an angle theta, take the sine, and add up the total generated at each stage to form a grand total.

[math]\sum\limits_{r = 0}^{r = n - 1} {\sin \left( {\theta  + r\alpha } \right)} [/math]

 

Edited September 1, 2017 by studiot

[Country Boy]

b+ c+ d+ e+ ...= (a+ b+ c+ d+ e+ ...)- a.

 For m> 1, sum_{i= m}^infinity  a_i is equal to \sum_{I= 1}^infiity a_i - \sum_{i= 1}^{m- 1} a_i.  The second sum is just a finite sum so is easy.

  In particular. sum_{i= 0}^infinity 1/2^i=  2 so sum_{i=1}^\infinity= 2- 1= 1, sum_{i= 2}^infinity 1/2^i= 2- 1- 1/2= 1/2, etc.

Edited September 26, 2017 by Country Boy