10 - Series and Sigma Summation Notation - Part 1 (Geometric Series & Infinite Series)

In this lesson, you will learn what a series is in math and how they compare to sequences. A series is the sum of the terms of a sequence. We have already discussed some special sequences such as the arithmetic sequence and geometric sequence. We now extend this idea and discuss the arithmetic and geometric series, infinite series, finite series, and more. We also spend a great deal of time introducing the sigma notation for summation, which is how we actually write the series down when solving problems..

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[ኢትዬ]

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[LarryZAR]

9:11
In order for you to get to 1 you have to get to 1/2 before that 1/4 before that 1/8 etc
Therefore
sum of (1/2+1/4+1/8...)=1

[rrrbb]

whenever possible, it's helpful to define infinite series using recursion. you can just solve for its value in most cases:

S = 1 + (1/2)S.
(1-(1/2))S = 1.
(1/2)S = 1.
S=2.

S= (1/2)^0 + (1/2)^1 + (1/2)^2 + ... + (1/2)^n + (1/2)^(n+1) S.
= 1 + 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + (1/2)^(n+1)S.

Note that the last term is multiplied times S. It's the tail of S expanded n times. But something AMAZING happens if you are careful. S isn't really the value. If the tail of the recursion goes to zero S = Sum(S, n), when n goes to infinity, because Tail(S, n) goes to zero.

S = Sum(S, n) + Tail(S, n).

This lets you calculate the closed form formula for n terms!

S - Tail(S, n) = Sum(S, n).

S - (1/2)^(n+1)S = Sum(S, n).
S(1 - (1/2)^(n+1)) = Sum(S, n).
2(1 - (1/2)^(n+1)) = Sum(S, n).
2 - (1/2)^n = Sum(S, n).

try it out...

adding the first 4 terms 0..3 replacements is...

Sum(S, 3) = 1/1 + 1/2 + 1/4 +1/8 = 15/8
= 2-(1/2)^3 = 2 - (1/8) = (2*8-1)/8 = 15/8

I honestly have no idea how people figure out most infinite series closed forms without using recursion.

Note the sum like 1/2 + 1/4 + 1/8 + ... = 1:

S = 1/2 + 1/2 S.
2S = 1 + S.
S = 1.
S
= 1/2 + 1/2 (1/2 + 1/2 S)
= 1/2 + 1/4 + 1/8 + ... + 1/(2^n) + 1/(2^(n+1))S.

and this to top it off:

S = 0.9 + 0.1 S
10 S = 9 + S
9 S = 9
S = 1

S = 1
= 0.9 + 0.1(0.9 + 0.1 S)
= 0.9 + 0.09 + 0.01(0.9 + 0.1 S)
= 0.9999....

And this works perfectly well for divergent series like "-1 = 1+2+4+8+...", where it is very clear what's going on. S=Sum(S, n) only when Tail(S, n) is zero; S is an important number in calculating the closed form; and is not necessarily the total. It is part of the recursive definition that replaces infinite iteration.

-1/12 = S
-1 = 12 S
1 = -12 S
1 = (1-13)S
13 S + 1 = S

S = 1 + 13 S
= 13^0 + 13 S
= 13^0 + 13(13^0 + 13 S)
= 13^0 + 13^1 + 13^2 + ... + 13^n + 13^(n+1) S

subtract the tail, and you have a formula for the first 13 powers.

more generally...

A = 1 + x A
(1-x)A = 1/(1-x)

When you differentiate this, you get the famous "-1/12 = 1+2+3+4+..." strange sequence. It is no paradox though because S isn't the value. S-Tail(S, n) is the value, and that goes to infinity as n goes to infinity, and it gives you the n(n+1)/2 formula, in case you didn't know it.

[dfuzesavetf]

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