18
$\begingroup$

Possible Duplicate:
What does the math notation $\sum$ mean?

My school's prescribed book uses the weird letter E character without explaining what it is in the first chapter when it talks about the binomial equation. I can't find it on Google either because I don't know what it means or its name. Please help me!

$$ (x+a)^n = \sum_{k=0}^{n} \binom{n}{k}x^ka^{n-k}$$

$\endgroup$
  • 10
    $\begingroup$ It's the Greek capital letter $\Sigma$ sigma. Roughly equivalent to our 'S'. It stands for 'sum'. Read this for starters. $\endgroup$ – Jyrki Lahtonen Nov 14 '11 at 7:27
  • $\begingroup$ en.wikipedia.org/wiki/Summation $\endgroup$ – Peđa Terzić Nov 14 '11 at 7:28
  • 1
    $\begingroup$ ...and maybe it wouldn't hurt to mention that sigma is a consonant, since called it a "weird E letter" could create a different impression. $\endgroup$ – Michael Hardy Nov 14 '11 at 12:04
  • $\begingroup$ In general $\displaystyle{\sum_{n=i}^{j} f(n) = f(i) + f(i+1) + \ldots + f(j-1) + f(j)}$ $\endgroup$ – badp Nov 14 '11 at 14:12
22
$\begingroup$

This is a capital sigma. Its use is best illustrated by an example:

$$ \sum_{k = 1}^4 \frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4}. $$

You begin by replacing the index (in this case, $k$) with the first value it takes on (in this case, 1). You then proceed to the next number and keep doing this replacement until you are at the upper limit (in this case, 4). Finally, you add all these terms up.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ It sort of reminds me of a for loop in programming a little bit. $\endgroup$ – Zell Faze Sep 27 '16 at 16:58
  • 1
    $\begingroup$ On the contrary, I tell my math students that the for loop should remind them of sigma notation. :) $\endgroup$ – Austin Mohr Sep 27 '16 at 17:24
6
$\begingroup$

As explained by Austin Mohr, it is the summation operation $\sum\limits_{k=m}^n f(k)$ (Greek uppercase letter Sigma) which sums every value of $f(k)$ where every value between $m$ and $n$ inclusively is substituted into $k$. That is:

$$\sum_{k=1}^5\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} \approx{2.28}$$

A related operation is the product operator denoted $\prod\limits_{k=m}^n f(k)$ (Greek uppercase letter Pi) which returns the product of the terms with $k$ substituted for every value between $m$ and $n$ inclusive.

$$\prod_{k=1}^3 (k + x) = (1 + x) (2 + x) (3 + x) = x^3 + 5x^2 + 11x + 6$$

| cite | improve this answer | |
$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.