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.