We can describe sums with multiple terms using the sigma operator, Σ. Learn how to evaluate sums written this way.

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s.v.jayachand

6 years agoPosted 6 years ago. Direct link to s.v.jayachand's post “can there be 2 sigma symb...”

can there be 2 sigma symbols at a time

and what if such condition occurs•

(43 votes)

kubleeka

6 years agoPosted 6 years ago. Direct link to kubleeka's post “Yes, there can. Something...”

Yes, there can. Something like Σ³ᵢ₌₁Σ⁴ⱼ₌₁ ij

In this case, we evaluate the innermost (rightmost) sum first. In the end, this will give us a function of i, which we then compute normally.

The inner sum is i+2i+3i+4i, or 10i. So this simplifies to Σ³ᵢ₌₁ 10i, or 10+20+30=60.

(130 votes)

G Y

7 years agoPosted 7 years ago. Direct link to G Y's post “In the first section (Unp...”

In the first section (Unpacking Sigma Notation), I've seen the index equal 0. But my calculus teacher says that the index can't be 0, because you can't have the 0th term of a sequence. But all else being equal (the sequence and summation index remaining the same), what would be the difference between a sum with i = 0 and a sum with i = 1?

Thank you.

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(11 votes)

andrewp18

7 years agoPosted 7 years ago. Direct link to andrewp18's post “Nothing really. Nothing c...”

Nothing really. Nothing changes if you shift all the indices down by 1. In fact, you can really start at any index you want because there's no convention that the subscript has to denote which number the term is in the sequence. Generally, people start at index 1 because it happens to be convenient to use the subscripts (and so the indices) to keep track of the number of the terms.

(17 votes)

Grave Watcher 69421

4 years agoPosted 4 years ago. Direct link to Grave Watcher 69421's post “How do i derive the formu...”

How do i derive the formula for summation?

Sum from k to n i = [(n-k+1)(n+k)]/2

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(7 votes)

Ian Pulizzotto

4 years agoPosted 4 years ago. Direct link to Ian Pulizzotto's post “Another way to derive thi...”

Another way to derive this formula is to let

S = Sum from k to n of i, write this sum in two ways, add the equations, and finally divide both sides by 2. We haveS = k + (k+1) + ... + (n-1) + n

S = n + (n-1) + ... + (k+1) + k.When we add these equations, we get 2S on the left side, and n-k+1 column sums that are each n+k on the right side.

So 2S = (n-k+1)(n+k).

Dividing both sides by 2 gives S = [(n-k+1)(n+k)]/2.(6 votes)

Cassie Areff

6 years agoPosted 6 years ago. Direct link to Cassie Areff's post “In my physics class the d...”

In my physics class the derivative of momentum was taken and the summation went from having k=1 on the bottom and N on the top to just k on the bottom, why is this? Is it the same thing, but short hand?

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(8 votes)

Ross Metcalf

5 years agoPosted 5 years ago. Direct link to Ross Metcalf's post “How do I solve for the nu...”

How do I solve for the number on top of the Sigma?

If I know the starting index, and I know the formula and the final sum, how do I solve for the ending index?

Example, how do I solve for X?

X

E f(n * 500) = 18000

n = 1•

(5 votes)

372447

3 years agoPosted 3 years ago. Direct link to 372447's post “Why is (sigma; n=1 to 3) ...”

Why is (sigma; n=1 to 3) (n!) -n undefined?

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(3 votes)

Hecretary Bird

3 years agoPosted 3 years ago. Direct link to Hecretary Bird's post “I have no idea, it doesn'...”

I have no idea, it doesn't look like it should be undefined. If we just calculate the sum from 1 to 3, we get a perfectly defined number:

Sum = (1! - 1) + (2! - 2) + (3! - 3)

= 0 + 0 + 3 = 3

If you set n to infinity though, the series will diverge and there will be no sum. Could you be confusing it with that?(4 votes)

brickbuilderchallenge

a year agoPosted a year ago. Direct link to brickbuilderchallenge's post “What would a function lik...”

What would a function like this look like in Khan Academy Javascript?

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(2 votes)

Cedar

a year agoPosted a year ago. Direct link to Cedar's post “Basic summation demo with...”

Basic summation demo with example in article:

`// Your function`

function f(x){ return 2*x - 1; }// Summation

function summation(lower, upper, f){

var sum = 0;

for(var i = lower; i <= upper; i++){

sum += f(i);

}

return sum;

}// Result gives 9, which is equal to what's expected.

// Arguments: lower bound on x, upper bound on x, f(x)

println(summation(1, 3, f));

You can modify the function 'f' in any way you want.(5 votes)

kaylebgable1453

a year agoPosted a year ago. Direct link to kaylebgable1453's post “This seems kind of sus”

This seems kind of sus

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(4 votes)

ZS

4 years agoPosted 4 years ago. Direct link to ZS's post “What if my index is m and...”

What if my index is m and is in the exponent of some "not index" variable?

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(2 votes)

Silentmists

4 years agoPosted 4 years ago. Direct link to Silentmists's post “Then you would still subs...”

Then you would still substitute the "m" values into the sum. The base variable would remain unchanged.

(4 votes)

Nick Voelckers

5 years agoPosted 5 years ago. Direct link to Nick Voelckers's post “Can the index only increa...”

Can the index only increase by 1, or is it possible for an index to increase by a larger number?

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(2 votes)

loumast17

4 years agoPosted 4 years ago. Direct link to loumast17's post “You can decide that. you...”

You can decide that. you give it a starting point and ending point, but you can use terms in the function to make it go faster or slower.

You could say the sum from i=0 to n of something with i^2. or you could even have something to the i power. But you will go through the integers with i, yes.So any manipulation of skipping numbers would have to be done in the function. For instance if you only wanted even numbers you would have a 2i in the function.

(3 votes)