Tutorial on Summation and #1 of DeNardo’s assignment on Least Squares Doron ZEILBERGER

Reminder about simple average If X_{1}, …, X_{n} are n numbers their average denoted by X̄ is defined by

X̄ = \frac{X_{1} + … + X_{n}}{n} ,

and in fancy sigma notation

X̄ = \frac{∑_{i=1}^{n} X_{i}}{n} .

An important consequence is that

∑_{i=1}^{n} X_{i} = nX̄ .

For the special case n = 2, we have

X_{1} + X_{2} = 2X̄ ,

and for the special case n = 3, we have

X_{1} + X_{2} + X_{3} = 3X̄ .

Reminder on Summation Given n numbers, a_{1}, a_{2}, … a_{n}, their sum, a_{1} + a_{2} + ... + a_{n} is abbreviated as

∑_{i=1}^{n} a_{i} .

There are simple rules on handling ∑ expressions. If c is any number, then

∑_{i=1}^{n} ca_{i} = c (∑_{i=1}^{n} a_{i})  (Rule1)

Indeed, when n = 2, for example, the left side is

ca_{1} + ca_{2}

simplifying, we get that this is

c(a_{1} + a_{2})

which is the same as the right side for n = 2.

Again, for n = 3. The right side is

ca_{1} + ca_{2} + ca_{3}

simplifying, we get that this is

c(a_{1} + a_{2} + a_{3})

which is the same as the right side for n = 3. Another obvious rule (in fact a special case) is

∑_{i=1}^{n} c = cn  (Rule)

Since

∑_{i=1}^{n} c = ∑_{i=1}^{n} c · 1 = c (∑_{i=1}^{n} 1) = cn ,

since ∑_{i=1}^{n} 1 = 1 + 1 + … + 1 (n times) and this equals n. Now I will spell-out Problem 1 for n = 2 and n = 3, but you would have to do it for general n using ∑ notation. When n=2 we have

w_{1} = \frac{X_{1} − X̄}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2}} , w_{2} = \frac{X_{2} − X̄}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2}} .

For the sake of simplicity let’s call the denominator (that is some number, and it is the same for w_{1} and w_{2}), D. In other words

D = (X_{1} − X̄)^{2} + (X_{2} − X̄)^{2} .

So abbreviating

w_{1} = \frac{X_{1} − X̄}{D} , w_{2} = \frac{X_{2} − X̄}{D} .

When n = 2 the three assertions spell-out to be

w_{1} + w_{2} = 0  (Assertion1)
w_{1}x_{1} + w_{2}x_{2} = 1  (Assertion2)
w_{1}^{2} + w_{2}^{2} = \frac{1}{D}  (Assertion3)

As for (Assertion1)

w_{1} + w_{2} = \frac{X_{1} − X̄}{D} + \frac{X_{2} − X̄}{D} = \frac{X_{1} − X̄ + X_{2} − X̄}{D} = \frac{X_{1} + X_{2} − 2X̄}{D} = \frac{0}{D} = 0 ,

since, as we said above (for n = 2) X_{1} + X_{2} = 2X̄, so X_{1} + X_{2} − 2X̄ = 0. For (Assertion2), it is better to use x_{i}, (rather than X_{i}), so we have, from the definition

w_{1} = \frac{x_{1}}{D} , w_{2} = \frac{x_{2}}{D} .

So

∑_{i=1}^{2} x_{i}w_{i} = x_{1}w_{1} + x_{2}w_{2} = x_{1} \frac{x_{1}}{D} + x_{2} \frac{x_{2}}{D} = \frac{x_{1}^{2}}{D} + \frac{x_{2}^{2}}{D} = \frac{x_{1}^{2} + x_{2}^{2}}{D} = 1 ,

since by definition, D = x_{1}^{2} + x_{2}^{2}. For (Assertion3), it is also better to use x_{i}

∑_{i=1}^{2} w_{i}^{2} = w_{1}^{2} + w_{2}^{2} = (\frac{x_{1}}{D} )^{2} + (\frac{x_{2}}{D} )^{2} = \frac{x_{1}^{2}}{D^{2}} + \frac{x_{2}^{2}}{D^{2}} = \frac{x_{1}^{2} + x_{2}^{2}}{D^{2}} .

But, by definition (when n = 2), D = x_{1}^{2} + x_{2}^{2}, so this is

\frac{D}{D^{2}} = \frac{1}{D} = \frac{1}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2}} .

This concludes the case n = 2. Now let’s do the same for n = 3.

w_{1} = \frac{X_{1} − X̄}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2} + +(X_{3} − X̄)^{2}} ,

w_{2} = \frac{X_{2} − X̄}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2} + +(X_{3} − X̄)^{2}} ,

w_{3} = \frac{X_{2} − X̄}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2} + +(X_{3} − X̄)^{2}} ,

For the sake of simplicity let’s call the denominator (that is some number, and it is the same for w_{1}, w_{2} and w_{3}), D. In other words

D = (X_{1} − X̄)^{2} + (X_{2} − X̄)^{2} + +(X_{3} − X̄)^{2} .

So abbreviating

w_{1} = \frac{X_{1} − X̄}{D} , w_{2} = \frac{X_{2} − X̄}{D} , w_{3} = \frac{X_{3} − X̄}{D} .

In terms of x_{i} rather than X_{i}, we have

w_{1} = \frac{x_{1}}{D} , w_{2} = \frac{x_{2}}{D} , w_{3} = \frac{x_{3}}{D} .

When n = 3 the three assertions spell-out to be

w_{1} + w_{2} + x_{3} = 0  (Assertion1)
w_{1}x_{1} + w_{2}x_{2} + w_{3}x_{3} = 1  (Assertion2)
w_{1}^{2} + w_{2}^{2} + w_{3}^{2} = \frac{1}{D}  (Assertion3)

As for (Assertion1)

w_{1} + w_{2} + w_{3} = \frac{X_{1} − X̄}{D} + \frac{X_{2} − X̄}{D} + \frac{X_{3} − X̄}{D} = \frac{X_{1} + X_{2} + X_{3} − 3X̄}{D} = \frac{0}{D} = 0 ,

since, as we said above (for n = 3) X_{1} + X_{2} + X_{3} = 3X̄, so X_{1} + X_{2} + X_{3} − 3X̄ = 0. For (Assertion2), it is better to use x_{i}, (rather than X_{i}), so we have, from the definition

w_{1} = \frac{x_{1}}{D} , w_{2} = \frac{x_{2}}{D} , w_{3} = \frac{x_{3}}{D} .

So

∑_{i=1}^{3} x_{i}w_{i} = x_{1}w_{1} + x_{2}w_{2} + x_{3}w_{3} = x_{1} \frac{x_{1}}{D} + x_{2} \frac{x_{2}}{D} + x_{3} \frac{x_{3}}{D} = \frac{x_{1}^{2}}{D} + \frac{x_{2}^{2}}{D} + \frac{x_{3}^{2}}{D} = \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}{D} = 1 ,

since by definition D = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}. For (Assertion3), it is also better to use x_{i}

∑_{i=1}^{3} w_{i}^{2} = w_{1}^{2} + w_{2}^{2} + w_{3}^{2} = (\frac{x_{1}}{D} )^{2} + (\frac{x_{2}}{D} )^{2} + (\frac{x_{3}}{D} )^{2}

= \frac{x_{1}^{2}}{D^{2}} + \frac{x_{2}^{2}}{D^{2}} + \frac{x_{3}^{2}}{D^{2}} = \frac{x_{1}^{2} + x_{2}^{2} + x_{3}^{2}}{D^{2}} .

But, by definition (when n = 3), D = x_{1}^{2} + x_{2}^{2} + x_{3}^{2}, so this is

\frac{D}{D^{2}} = \frac{1}{D} = \frac{1}{(X_{1} − X̄)^{2} + (X_{2} − X̄)^{2} + (X_{3} − X̄)^{2}} .

Now it is your job to do it for general n using ∑ notation!

Doron Zeilberger, Department of Mathematics, Rutgers University (New Brunswick), Hill Center-Busch Campus, 110 Frelinghuysen Rd., Piscataway, NJ 08854-8019, USA.

Email: zeilberg at math dot rutgers dot edu .