www.mathcentre.ac.uk/.../Sigma%20notation.mp4

This video is all about Cigna
notation. This is a concise and

precise way of explaining long

sums. We study some examples and
some special cases and then will

drive some general results.

But first, let us look at these.

In math we have to, sometimes

some. A number of terms in the
sequence, like these two.

The first one, 1 + 2 + 3 +
4 + 5. It's the sum of the

first five whole numbers.

The second one, 1 + 4 + 9 +
16 + 25 + 36. It's the sum of

this first 6 square numbers.

Now, if we had a general
sequence of numbers such

as this.

You want you to you three, and
so on. We could write the sum as

Air Sequel you one, plus you 2
plus you three an so off. Now

this some goes on and on and on.
But if we want to some and terms

then we can write that the sum
for end terms equals you one

plus you too.

Bless you 3 up to

UN. And those are in terms.

Now we can use the Sigma
notation to write this more

concisely. And Sigma comes from
the Greek Capital Letter, which

corresponds to the S in some.
And it looks like this Sigma.

And if we want to express this
some using Sigma notation, we

take the general term you are.

And we say that we are summing
all terms like you are from R

equal 1 two N.

And that equals are some SN
which is written above.

More generally, we might take
values of our starting at any

point rather than just at one
and finishing it in so we can

write Sigma from our equal A to
be have you are and that means

the sum of all the terms.

Like you are starting with our
equal, a two are equal, BA is

the lower limit.

And B is the upper limit.

Now what I'd like to do
is explore some examples

using the Sigma Notation.

So take for
example this one.

It's Sigma of R cubed.

With our starting are equal 1
and ending with R equals 4. So

the lower limit is are equal 1
and the upper limit is are

equal. 4 will all we have to do
is just expanded out and write

it as along some using the
values are equal 123 and ending

at our equal 4.

When RS1R cubed is 1 cubed,
are equal 2 two cubed are

equal 3 three cubed and are
equal 4 four cubed.

And we can just.

Work the Zeit and add them
together. 1 cubed, 12 cubed is

it 3 cubed? Is 3 * 3 *
3 so that's 936-2074 cubed is 4

* 4 which is 16 times by 4
which is 64 and when we Add all

of those together you get a
lovely answer 100.

What about this one? It's
similar, but instead of using R

is the variable. You can use N.
In fact, you can use any

variable that you want.

And in this case were summing
terms like N squared for men

starting from 2 to 5, going up
in increments of one.

So starting with an equal 2.

And squared is 2 squared.

Plus

An equal 3 three squared. An
equal 4 four squared and

finishing with an equal 55

squared. And all we have to do
now is square these out and add

them together 2 squares for
three squared, 9 four squared 16

and 5 squared is 25.

And we can add these very
simply. Foreign 16 is 20.

9 and 25 is 3420
and 34 is 54.

What could be simpler?

Look at these two written using
the Sigma notation, just

slightly different from the
previous ones. Remember we look

at all the terms.

Such as two to the K and we
substitute in the values for K

in increments of one starting
with zero going up to five.

So the first one 2 to the K when
K is 0, is 2 to the North. Then

we add 2 to the one +2 to the
2 + 2 to 3 + 2 to the

4 + 2 to the five.

Notice we've got six terms and
the reason why we've got six

terms is we started K at zero
and not at one.

And then we evaluate those two
to the zero 1 + 2 plus

force plus three 2 cubed, which
is it +2 to 416. Let's do

the 5:32. And when we
add those up, 1 + 2

is 347-1531.
And then 63.

Now look at this one.

A little bit more involved, but
quite straightforward if you

follow the same steps that we've
done before, we take the general

term here and we substitute in
values looking at where it

starts from and where it ends.
So R equals one is the lower

limit, so we look at our general
term and substituted are equal 1

first and that will give us one
times by two.

Times by 1/2 added

two. Now we substitute are equal
to for the next one, so that's

two are plus one is 3 +
1/2. An hour equals 3.

3 plus One is 4.

Plus 1/2 an.

And I forgot my arm, but my are
in this case is 4 and we add 1

to it which is 5 and continue on
in the same then until we get to

our upper limit our equals 6. So
that's a half six times by 7.

And we can evaluate this side,
but we can do a lot of

canceling on the way before we
do. And if the adding two

cancers with the two here and
here, 2 cancers with the four

to give us two here 2 cancers
with this forward to give us 2

two counters with this six to
give us three and two counters

with this six to give us 3.

And then we evaluate eyes adding
together terms. So this one is 1

plus. And we're left with 1
* 1 * 3, which is 3

+ 1 * 3 * 2, which
is 6 plus.

2 * 5 which is 10.

Plus 3 * 5 which is 15 and
the last 121.

Now if you look at these
numbers, these are quite

special numbers. There's a
name given to them, they

either triangular numbers.

And you should be able to
recognize these the same way as

you would recognize square
numbers or prime numbers, or

order even numbers. I'm finally
what would like to do is just

add these altogether. 1 + 3 + 6
is 10 + 10 is 20.

20 + 1535 thirty six and then
add the 20 is 56.

So we get to some 56.

Now, what

about this?
We've got our general term 2 to

the K and we want to start with
K equal 1 decay equal an if we

write it out substituting in our
values for K starting with K

equal 1. Our first time will be
2 to the one plus then case two

would be 2 to the 2 + 2
to 3 + 2 to the 4th and

so on. Until we get to
two to the N.

Now we can't evaluate
this to a finite number,

because at this stage we
don't know what Aaron is.

And what if we
have these two expressions?

But still using the Sigma
notation, but in this case are

general term involves a negative
sign in both cases.

So we'll just take a little bit
of time and carefully work the

might. Because it's dead easy to
go wrong in these sorts of ones,

so take a little bit of time and
you will hopefully work them out

OK using the same steps that

we've done before. In this
particular example, our general

term is negative, one raised to
the part of our and R equals 1

to 4. So the first term will be
negative. One reason the part of

one. Plus negative one race to
the power of 2 plus negative

one. Raise the pirate three and
finally negative one raised to

the power of 4.

And then all we have to do is
multiply these item, making sure

that we get our signs correct,
negative one raise the part of

one is negative, one negative
one raised to the part of two

that's squared is positive one.

Negative 1 cubed.

That will give negative one and
negative one is part of four.

That'll give positive one.

So in fact.

Are some work site to be O
negative 1 + 1 zero plus

negative 1 + 1 zero?

Quite a surprise, in fact, in

fact. Now look at this one.

Again, take care. Our general
term is negative one over K

squared and we have to use
values of K from one to three.

So we start with K equal
1. That's negative one over 1

squared plus negative one over 2
squared plus negative one over 3

squared. And then again, taking
care with the negative signs.

Square these values out negative
one over one is negative, 1

squared will give us one plus.

Negative one over 2 all squared
that would be positive and is

1/2 * 1/2, which is quarter.

Then added two negative one over
3 squared. Again, that's going

to be positive, and it's going
to be one over 3 * 1 over 3,

which is one of her and I.

We can add these together,
adding the two fractions

together.

The common denominators 36.

And it will be 9 + 4 and
you've got your one there. So my

final answer is one and 13 over
36. Quite a grotty answer, but.

We've been able to workout what
seemed to be quite a complicated

Sigma notation some.

And we finally got an answer.

But what if we had along some
and we wanted to write it in

Sigma Notation? How would we do

it? Well, why don't we first
look at the two sums that we

started off with?

The first one was 1
+ 2 + 3 +

4 + 5.

This one is quite an easy one to
start off with because it's the

sum. What type of terms? Well,
it's a dead easy term because

they are adding one on each
time. So if we take the general

term to BK.

K starts the lower limit one and
ends with the upper limit 5.

So there's no problem with this
one. This is very

straightforward. But look at the

second one. Again, it's not too
bad because we notice and we

said before that these are

square numbers. So we
could actually write these

as one squared +2 squared,
+3 squared, +4 squared, 5

squared, and six squared.

Once we've done this, are Sigma

notation some? Is dead easy
because we could easily see our

general term must be K squared.

And K must start with cake, will

one. Takei equals 6.

But what
about these

ones? This is
a long some.

What's different about this?
It's got fractions and it also

has alternating signs.

Negative. Positive negative
positive negative positive

silver. Now the trick in this

case. Is to rewrite all of these
in terms of fractions.

And then use negative 1 to a

power. So if we look at this, we
can rewrite this as negative one

over 1. Plus 1/2.

Minus one over 3 +

1/4 plus. And So what to
one over 100?

Now we have to deal with
the signs alternating.

And if we think back we did have
signs alternative before.

And that's to do with negative
numbers. Now we don't want to

change the value of the
fraction, we just want to change

the sign. So if I rewrite
this as negative one.

Times by one over 1.

Plus Negative one.

The power of 2.

One over 2.

Plus negative 1 to the power of
3 one over 3.

Plus negative 1 to the power of
4 one over 4.

Time 2 plus negative one to
the 101 over 100.

Have we got the same sum?

One yes, because
look at each term.

Negative 1 * 1 over one is
negative, one over 1.

Negative 1 squared will be
positive 1 * 1/2, which will

give us our plus 1/2.

Negative 1 cubed will be
negative one times by 1/3 which

is negative 1/3, so that's where
we get the negative 1/3 negative

1 to the power of 4.

Is positive one times by 1/4
gives us our plus one over 4.

So rewriting our sequence that
we started off with here.

Into this format.

That then helps us to write it
in Sigma Notation.

Because we can easily see that
this is Sigma.

We know it's negative one.

To the power and the power
is always the number underneath

the fraction. So we'll say that
our fraction is one over K

because we start with K equal 1
and we raise negative 1 to the

power of K.

And to complete our Sigma
notation for this long some.

We start with the lower limit of
K equal 1.

And the upper limit K equal 100.

So you can easily see that this

notation. Is very concise.

And I really lovely way of
writing that big long some.

Now I'd like to move
on some special cases using

Sigma notation. The first
special case is when you have to

sum a constant. Like this?

If we have Sigma of three
from K equal 1 to 5, what

does that actually mean?

Well, we work at night.

We take each term for cake with
one up to K equal 5.

Each time remains the same. The
matter whether cake was one K

equals 2 cakes, 3, four, or
five. So we actually just get

the sum of 5 threes.

Which is 15th.

So with five terms, five times
the constant to make 15.

So we can get a general result
from that if instead of three we

have a constant C.

And we take our
K from one to N.

What does this work? I to be?
Well, we do it in the same way.

We workout are some using cake

with one. 2, three, and so on up
to K equal an.

And evaluate it.

So can equal 1 is say K
Equal 2C and so on.

Up to. The last see
such that we have and terms.

Because we've got N lots of see
this, some can be simplified

dead easily to an times C.

So this is a general result
which is very useful when using

Sigma notation. If you have to
sum a constant.

From cable one to N, then the
value is and it is a finite

value and times by C.

What could be easier?

But what about this?

What if you had to sum
a variable like this 3K?

Can this be written more easily?

Well, we'll take this particular

example. 3K some 4K. Will
want to four.

So will sub in cake. Will 1
first, then K equal 2 then K

equals 3 and then K equals 4 and
we get that some.

We notice that three is a factor
if each one of these terms.

So we factorize IR 3 and we
get three bracket 1 + 2 +

3 + 4.

Now that is equal to three times
by. If you want to add these

altogether, that's three 610. So
the answer is actually 30.

Now I'm not so much interested
in the answer, but in the

process. At anyone stage, can
you spot something that we know

already? Well, look at this
second line. You've got 3 times

by 1 + 2 + 3 + 4.

That's the addition of K&K is
one 2, three and four, and we

can rewrite this.

Using Sigma notation that

is 3.

Sigma. Values
K from K equals

124.

So our expression that we
started off with Sigma of three

K equals 1 to 4K was one
to four equals 3 times.

Sigma of K from cable one to
four. So effectively we've taken

out with factorized at three
from our first expression, and

we brought it outside the Sigma

notation. So more generally.

We can take that if we've got
the Sigma of CK from K equals

1 to N.

What would that equal?

Wow, quite easy. All we have to
do is that we see that that is.

See times one plus C times
2 plus C times 3.

Right up to.

See times an. We've got M terms

here. We can factorize by the

sea. And we get 1
+ 2 + 3 plus up

to N and then we can
rewrite this using Sigma

notation, that is C Sigma
of care from K equals 1

to N.

And as I say.

Quite straightforward, when
you've got a Sigma of C times a

variable. See being a constant,
you can take the constant

outside the Sigma notation and
you left with C times by the

Sigma of K from K equal 1 to

N. That's another very useful
result that you should remember

when using Sigma notation.

But what if we had this
expression? The sum of the

variable K plus two where K
takes the lower limit one to

four? What does that work? I'd
be? Can we write it more concert

concisely and easily well?

We do it stage by stage, step by
step, putting in values for K

from K equal 1 to 4.

So the first time case one says
1 + 2.

Plus second, her K equals 2, so
that's 2 + 2.

The third term is 3 + 2 and
the fourth term is 4 + 2.

Now what I want to do next is
rearrange these values.

If I rearrange such that I have
the addition of 1 + 2 + 3 + 4

first. And then see what
we've got left. We've got

add 2 four times, so
that is 4 * 2.

And then we can easily evaluate
this expression 1 + 2 + 3 + 4

is 10 + 4 two 38. So the answer

is 18. But as I said before,
we're not so much concerned with

the answer, but in the process,
can we look and see at any line?

Can we use the Sigma notation

more easily? Well, if we look at

this line. We've seen part
of this before.

1 + 2 + 3 + 4.

Can be written more easily using
Sigma notation as Sigma. Care

from K equals 1 to 4.

Plus then we've got 4
* 2.

So looking back at what we
started with, we've got Sigma of

K plus two from K equals one to
four and it expanded I to this

Sigma notation Sigma of can at
all, from cable one to 4 + 4

* 2 or two is the constant that
was added to the care.

So we can generalize
this result.

If we had Sigma
of.

G of K function of K
plus a constant C.

As our variable and we want to
take care from one to N.

We can use the pattern that
we spotted here to write this

out using Sigma notation plus
some other value.

The first part is Sigma.

Off the function that we started
off with in our variable and the

function is GF care.

And we see that it should be
from K to the upper limit from K

equal 1. Takei equal the upper
limit in this case is for, but

in our general case it's K equal

N. Says K equal 1 to an.

Plus Now the next part, remember
it was the upper limit by the

number of terms times by the
constant that you've had in your

variable. Well, look here. Are

constantly see. And our
upper variable is an.

So it's End Times C.

So this Sigma notation can

be expanded. Two Sigma GF
K from K Equal 1M Plus M

Times C.

And we can even generalize
further using previous results

if we took Sigma from K equals
1 to N.

Of a GF K

Plus C. As our

variable. We can expand that

out. Looking at this time 1st
and then dealing with the

constant now, this term is a
multiple a constant times by

your GF K so we can bring that
outside the Sigma notation.

And say it's a Times by
Sigma GfK from K equals 1

to N.

Plus

Just what's left to do with the
sea and we know when that's

inside the bracket when we
expand it out.

We're left with end times C.

What could be easier?

All right? But

what if? We wanted to
use Sigma notation and the

variable we wanted to add
together is K Plus K squared IE

two functions of K.

What would we get when we write
out this long some?

Start simply by substituting in

for K. Case one to begin with.
So the first term is 1 + 1

squared. Second time is 2 +
2 squared. Third time is 3 + 3

squared, and that's our final
term because the upper limit for

K is 3.

Now we'll rearrange the way we
did before to make our work

easier. Will take the 1 + 2
+ 3 together.

And then we'll take the one
squared and the two squared

and the three squared bits
that are left.

And you can see why I've done
that, because here we're adding

our constant case and here we're

adding. Our case squareds.

When we add these together, we
just add them up and multiply

out using the square sign you
get 1 + 2 + 3 is 6,

one squared is 1, two squared is
4 three squared 9. Add back

together that gives you
fourteen. 6 + 14 is 20.

But look at this second line.

We can simplify this second line
using Sigma notation.

And it's using Sigma notation
that we've done before. 1 + 2 +

3 is the same as Sigma of care
from K equals 1 to 3.

Plus

1 squared, +2 squared, +3
squared, that's dead easy. It's

the sum of all variables, K
squared from K equals 1 to 3.

Now look at what would started
and look what we've got.

We have broken this Sigma
notation up into two sums to

Sigma sums. So here we've
got the Sigma offer variable

plus another variable.

And with split it up to.

Sigma of one variable added to
the Sigma of the other variable.

We've used the distributive law.

So to write this more

generally. We can say that the
sum of a function of care.

Added to another
function of K.

From K equals 1 to N must
equal the sum.

Two separate sums. The sum of
JFK from K equals 1 to M

Plus Sigma. FFK from K
equals 1 to add.

And in fact, we can continue
this on if we added on other

functions within our variable

here. Then all we have to do
is add on another Sigma of

that particular function.

Finally, here's a real life
example using Sigma notation.

Take for example if we want
to find the mean of a set of

numbers. That could be the
marks in a test.

If we want to find the main what
we want to do is we have to

workout the total sum.

Divided by the number of values.

And that's what the main is.
Well, that's actually what we've

been doing in the previous
examples. We have been finding

the total sum, and we have been
looking at the number of values.

So if I take this example, just
say we had the marks 23456.

And we want to find the mean of
those marks. The main will

equal. 2 + 3 + 4 +
5 + 6 and we'd have to divide

it by the number of values,
which is 5 when we work that

out. That is 5 + 510, another

10. 20 over 5 which works out to
be the value 4.

But more generally.

If you've got a set of marks,
say XI, we can write the main

in terms of Sigma Notation. It's
the total sum of all the marks.

Such as Zhao.

From I equals 1 to N.

And then we want to divide it by
the number of values. Now we

know there are N Marks.

So we want to pie by one over
N, so the main is equal to one

over an Sigma X of I from I
equal 1 to M.