# Summation, Sigma

1. Feb 2, 2007

### disregardthat

(This is not a homework question!)

I have no education in this kind of math yet, but I wonder how many ways you are allowed to use the summation sign sigma. I can't seem to get a good explanation on google or wikipedia.

Since I like to try myself with tex, I will write an example of it here:

$$\sum_{k=3}^4 k^2$$ This will be a normal summation sign.

To see if I have got it right: $$k = 3$$ means that the k on the side of the sigma will start at 3, right?

If the sigma is raised to 4 like the one I have shown, it means that the k (3) will be added to a number 3+1, and then to 4+1, and then to 5+1. That the k is raised to the power of two, means that for each part of the serie, the number will be raised, like (3)^2 + (3+1)^2 + (4+1)^2... right?

I think this will be the same as: $$\sum_{k=3}^4 k^2 = 3^2 + 4^2 + 5^2 + 6^2 = 86$$

Is this the correct use of the Summation?

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I have heard that it is used in biology to find out the number of cells that is being reproduced.
Let's say that we have a cell, and it has unlimited food, so it will reproduce in the rate of doubling each ten minutes. And we will watch it for one minute. That means that the cell and it's daughter cells will reproduce 60\10 = 6 times.
I found out that the only way that can be done is if you put it up like this:

$$\sum_{k=0}^6 2^k$$

So I guess it would give us the right answer of how many cells that would be there.

$$\sum_{k=0}^6 2^k = 2^0 + 2^1 + 2^3 + 2^4 + 2^5 + 2^6 = 127$$

The amount of cells that we will end up with, assuming none of them died, and assuming every cell reproduced itself each ten minutes, after one minute.

Is this a valid way of using the summation? If not, it should be

Last edited: Feb 2, 2007
2. Feb 2, 2007

### cristo

Staff Emeritus
No; In words, $$\sum_{k=3}^4 k^2$$ means the "sum from k=3 to k=4 of k squared." So, this would be equal to 32+42.

This, however, is a correct use of the summation.

Note that the sigma notation is simply a shorthand way to write a long sum. To take a simple example, the sum 1+2+3+4+5+6 can be written in sigma notation as $$\sum_{n=1}^6n$$

Last edited: Feb 2, 2007
3. Feb 2, 2007

### disregardthat

Thanks! I see that the power you raise the sum to is the point where it stops, and not the number of times you add the number in. I understand it.

Is it possible to make the amount of increasment in each addition to be lower or higherthan 1? Example: Is it correct to use this kind if summation:
$$\sum_{k=2}^2 \frac {4}{k}$$ And that would be: 2/4 + 3/4 + 4/4 = 2.25

EDIT: I meant this

$$\sum_{k=2}^4 \frac {k}{4}$$ And that would be: 2/4 + 3/4 + 4/4 = 2.25

Or is another way to make the amount of increasment lower, higher or in different numbers, for example $$pi$$

Is this allowed for example, single yes and no answer is fine:

$$\sum_{k= \pi}^3k$$

Last edited: Feb 2, 2007
4. Feb 2, 2007

### cristo

Staff Emeritus
I'm not sure if I understand your question: That that use of the notation is fine, it would expand as $$\sum_{k=2}^8 \frac {4}{k} =\frac{4}{2} +\frac{4}{3}+\frac{4}{4}+\frac{4}{5}+\frac{4}{6}+\frac{4}{7}+\frac{4}{8}$$

5. Feb 2, 2007

### disregardthat

Sorry, I mispelled the sum up there. But I got my answer. Thanks.

But not on the last one. I wonder if it is possible to use $$\pi$$ instead of 1 as the rising on each addition. I guess it would stand like this:

$$\sum_{k= \pi}^3k$$

which would equal: $$= \pi + 2\pi + 3\pi = 6\pi$$

Last edited: Feb 2, 2007
6. Feb 2, 2007

### arildno

Why not just write that one as:
$$\sum_{k=1}^{3}\pi{k}$$

7. Feb 2, 2007

### disregardthat

Thanks

And how excactly do I find this on my calculator, sorry for asking this much, but we are not learning this at school...

I find " Sum( " on my TI-84 Plus calculator, and I read that it should be put up like this: $$\sum$$(K,K,1,10) (for example) But I don't know how.

8. Feb 2, 2007

### arildno

How should I know how your calculator works??

9. Feb 2, 2007

### cristo

Staff Emeritus
This sum $$\sum_{k=2}^4 \frac {4}{k}$$ does not equal 2/4 + 3/4 + 4/4, it equal 4/2 + 4/3 + 4/4.

I think you may be a little confused here. The limits on the sum (here k=2,...,4) means the first term in the sum has k=2, and then the next terms increase the value of k by 1 each time.

The summation notation means that we sum over integers k, starting with k=the lower limit, and ending with k= upper limit. It does not make sense to set the bottom limit to pi (since this is not an integer). If you wish to write the sum $\pi + 2\pi + 3\pi = 6\pi$ using sigma notation, then use arildno's suggestion above.

10. Feb 2, 2007

### disregardthat

I thought everyone got that calc, universal calc or something... Nevermind then, I'll sort it out somehow

Anyway, I understand it now, thank you.