ITEC122 2008fall-ibarland

Summations

Ian Barland

Rosen p.229–232 has a fine introduction to summations. The key element to remember is that ∑ is just a notation; whenever you see it you should mentally expand it into the sum it represents.

Here are a few problems from Rosen that introduce some standard tricks:

• Rosen 3.2, #15a:

∑_{j=0}^{8} 3 · 2^{j} = 3 · 2^{0} + 3 · 2^{1} + 3 · 2^{2} + ⋯ + 3 · 2^{8}
= 3(2^{0} + 2^{1} + 2^{2} + ⋯ + 2^{8})
= 3 · ∑_{j=0}^{8} 2^{j}
= 3 · (2^{9} − 1)by Rosen 3.2 Th’m 1

The handy trick is that you can pull out the constant factor 3.

• Rosen 3.2, #17d: The previous trick is often useful in double sums:

∑_{i=0}^{2} ∑_{j=0}^{3} ij = ∑_{i=0}^{2} (∑_{j=0}^{3} ij)

= ∑_{i=0}^{2} (i ∑_{j=0}^{3} j)

= ∑_{i=0}^{2} (i · 6)

= 6 ∑_{i=0}^{2} i
= 6 · 3
= 18

Why was it valid, in the first line, to factor out i from the inner sum? Because (with respect to the inner sum over j) it was a constant. Again, writing it out explicitly makes this clear.

• Difference of sums: When a sum’s initial index isn’t a nice even 0 or 1, often we can express the sum as a difference of two others. See Rosen Section 3.2, Example 15:

∑_{k=50}^{100} k^{2} = ∑_{k=1}^{100} k^{2} − ∑_{k=1}^{49} k^{2}

, and now each of the two sums can be individually computed from Section 3.2, Table 2.

• ∑_{i=4}^{7}(2i − 6) = 2 + 4 + 6 + 8 = 20. This sum notation can also be written ∑_{i∈{4,5,6,7}}. or ∑_{i∈[4,7]}.

Note: there’s an easier summation notation for 2+4+6+8: If we let j = i − 3, then when i = 4 then j = 1; and when i = 7 then j = 4; thus

∑_{i=4}^{7}(2i − 6) = ∑_{j=1}^{4} 2j

. (Or even start the sum from 0 :-)

• Know this sum cold:

∑_{i=0}^{n} i = n(n + 1)/2

. [the triangular numbers]

• Know this sum cold as well:

∑_{i=0}^{n} 2^{i} = 2^{n+1} − 1

. [examples for n=2,3]

• This previous example generalizes:

∑_{i=0}^{n} b^{i} = (b^{n+1} − 1)/(b − 1)

. [consider b = 10 : 1 + 10 + 100 + 1000 = 9999/9 = 1111 ]

When b < 1, we can take the infinite sum. In particular, 1 + 1/2 + 1/4 + +... = 2. 1 + 1/10 + 1/100 + ... = 1.11111 = 10/9. Note that .9 + .9^{2} + ... will also converge (to what?).

Similarly: Is .432432432432... rational? = 0.432∗∑_{i=0}^{∞} 1/1000^{i} = 0.432(−1/(−999/1000)) = 432/999. Indeed, this generalizes: any repeating-decimal is rational.

• Now try: ∑_{i=51}^{100} 2^{i}. Split into two different sums; subtract. Actually, it’s a bit moot for 2^{i}, because the first 50 terms altogether weren’t as big as the 51st, and the 51-54rd are nearly sixteen times as big.

• Double sums:

∑_{i} ∑_{j} i ∑_{i} ∑_{j} ij ∑_{i} ∑_{j} j^{i} What if we switch order? To find out, expand!

• Consider expectations:

the ”mean” value of a fair six-sided die is ∑ i·; (1/6). For an n-sided die, ∑ i/n. How about for a weighted 6-sided die, where the two-pips has been artfully changed into a three-pips side:

Expected number of tosses until a coin(die) comes up heads(6)? ... Can expand as rows and columns; arrange creatively and re-add. [Okay, it’s a bit fishy w/ infinite series, but we’ll hush that up.]