How to interpret this Sigma notation.

[kieth89]

Homework Statement

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.

Homework Equations

None

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

 

[AmritpalS]

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

 

[Bread18]

$a_i$ 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 $a_0 = 1, a_1 = 2, a_3 = 3, a_4 = 2, a_1 = 1$

 

[HallsofIvy]

$n= \sum_{i= 0}^k a_ib^i$
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, $a_0= 2$, $a_1= 2$, $a_2= 1$, $a_3= 2$, $a_4= 0$, $a_5= 1$.

 

[kieth89]

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

 

[Ray Vickson]

Homework Statement

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.

Homework Equations

None

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

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

RGV

 

[dimension10]

Homework Statement

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.

Homework Equations

None

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

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,

[tex]\sum_{x=0}^{\infty}f(x)[/tex]

results in

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

Alternatively, you could write

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

And for itex codes you can also write:

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

which results in:

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