# How to interpret this Sigma notation.

1. Jun 19, 2012

### kieth89

1. The problem statement, all variables and given/known data
Taken from Wikipedia's Palindromic number entry, under formal definition header:

Although palindromic numbers are most often considered in the decimal system, the concept of palindromicity can be applied to the natural numbers in any numeral system. Consider a number n > 0 in base b ≥ 2, where it is written in standard notation with k+1 digits ai as:

$n = ^{k} _{i=0}\sum{a_{i}b^{i}}$
NOTE: The K should be above the sigma, the i = 0 below..but I couldn't figure out the Itex notation for that..

with, as usual, 0 ≤ ai < b for all i and ak ≠ 0. Then n is palindromic if and only if ai = ak−i for all i. Zero is written 0 in any base and is also palindromic by definition.

2. Relevant equations
None

3. The attempt at a solution
Can't really attempt it..basically I just don't know how to interpret the a_sub i part..what does that mean? Is this showing me a pattern that I can use to determine if a number is palndromic? If so, what is the pattern.

sub letters have always bugged me.., thanks for any help

2. Jun 19, 2012

### AmritpalS

A sub i, i believe, is telling you which term in the sequence

3. Jun 19, 2012

$a_i$ is just some natural number. For n to be palindromic, $a_i = a_{k-i}$

For example, with the number n = 12321, we have n = $\sum^{k}_{i=0}a_ib^i$. n has 5 digits, so k = 4, and it's a base 10 number, so b = 10. so we have
$n=\sum^{4}_{i=0}a_i10^i = a_0 10^0 +a_1 10^1 +a_2 10^2 + a_3 10^3 + a_4 10^4$

For this number we take $a_0 = 1, a_1 = 2, a_3 = 3, a_4 = 2, a_1 = 1$

4. Jun 19, 2012

### HallsofIvy

Staff Emeritus
$n= \sum_{i= 0}^k a_ib^i$
just says that a number written in base b is a sum of numbers, $a_i$, the "digits", times powers of the base. The number 12034, in base 10 means 1(10000)+ 2(1000)+ 0(100)+ 3(10)+ 4(1)= 1(10^4)+ 2(10^3)+ 0(10^2)+ 3(10^1)+ 4(10^0). The number 314 can be exanded in powers of 3, say. 3^2= 9, 3^3= 27, 3^4= 81, 3^5= 243, and 3^6= 729 which is larger than 314. 243 divides into 314 once with remainder 71. That is, 314= 1(3^5)+ 71. 81 does not divide into 71 so 314= 1(3^5)+ 0(3^4)+ 71. 27 divides into 71 twice with remainder 17 so 314= 1(3^5)+ 0(3^4)+ 2(3^3)+ 17. 9 divides into 17 once with remainder 8: 314= 1(3^5)+ 0(3^4)+ 2(3^3)+ 1(3^2)+ 8. Finally, 3 divides into 8 twice with remainder 2 so 314= 1(3^5)+ 0(3^4)+ 2(3^3)+ 1(3^2)+ 2(3^1)+ 2(3^0). That is saying precisely that $314= \sum_{i=0}^5 a_i b^i$ with b= 3, $a_0= 2$, $a_1= 2$, $a_2= 1$, $a_3= 2$, $a_4= 0$, $a_5= 1$.

5. Jun 19, 2012

### kieth89

Thank you all for the very speedy and very thorough breakdowns. It makes much more sense now.

6. Jun 19, 2012

### Ray Vickson

To get $\sum_{i=0}^k ...$ use "[ i t e x] \sum_{i=0}^k ... [/i t e x]" (no spaces).

RGV

7. Jun 21, 2012

### dimension10

What are you trying to show?

P.S. In order to put the index and stuff above/below the sigma use tex instead of itex. For example,

Code (Text):
$$\sum_{x=0}^{\infty}f(x)$$
results in

$$\sum_{x=0}^{\infty}f(x)$$

Alternatively, you could write

Code (Text):
$$\sum_{x=0}^{\infty}f(x)$$
And for itex codes you can also write:

Code (Text):
$\sum_{x=0}^{\infty}f(x)$
which results in:

$\sum_{x=0}^{\infty}f(x)$