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

• 0:02 - 0:09
This video is all about Cigna
notation. This is a concise and
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precise way of explaining long
• 0:11 - 0:17
sums. We study some examples and
some special cases and then will
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drive some general results.
• 0:20 - 0:23
But first, let us look at these.
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In math we have to, sometimes
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some. A number of terms in the
sequence, like these two.
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The first one, 1 + 2 + 3 +
4 + 5. It's the sum of the
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first five whole numbers.
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The second one, 1 + 4 + 9 +
16 + 25 + 36. It's the sum of
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this first 6 square numbers.
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Now, if we had a general
sequence of numbers such
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as this.
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You want you to you three, and
so on. We could write the sum as
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Air Sequel you one, plus you 2
plus you three an so off. Now
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this some goes on and on and on.
But if we want to some and terms
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then we can write that the sum
for end terms equals you one
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plus you too.
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Bless you 3 up to
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UN. And those are in terms.
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Now we can use the Sigma
notation to write this more
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concisely. And Sigma comes from
the Greek Capital Letter, which
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corresponds to the S in some.
And it looks like this Sigma.
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And if we want to express this
some using Sigma notation, we
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take the general term you are.
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And we say that we are summing
all terms like you are from R
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equal 1 two N.
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And that equals are some SN
which is written above.
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More generally, we might take
values of our starting at any
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point rather than just at one
and finishing it in so we can
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write Sigma from our equal A to
be have you are and that means
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the sum of all the terms.
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Like you are starting with our
equal, a two are equal, BA is
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the lower limit.
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And B is the upper limit.
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Now what I'd like to do
is explore some examples
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using the Sigma Notation.
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So take for
example this one.
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It's Sigma of R cubed.
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With our starting are equal 1
and ending with R equals 4. So
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the lower limit is are equal 1
and the upper limit is are
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equal. 4 will all we have to do
is just expanded out and write
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it as along some using the
values are equal 123 and ending
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at our equal 4.
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When RS1R cubed is 1 cubed,
are equal 2 two cubed are
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equal 3 three cubed and are
equal 4 four cubed.
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And we can just.
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Work the Zeit and add them
together. 1 cubed, 12 cubed is
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it 3 cubed? Is 3 * 3 *
3 so that's 936-2074 cubed is 4
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* 4 which is 16 times by 4
which is 64 and when we Add all
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of those together you get a
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similar, but instead of using R
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is the variable. You can use N.
In fact, you can use any
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variable that you want.
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And in this case were summing
terms like N squared for men
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starting from 2 to 5, going up
in increments of one.
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So starting with an equal 2.
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And squared is 2 squared.
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Plus
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An equal 3 three squared. An
equal 4 four squared and
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finishing with an equal 55
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squared. And all we have to do
now is square these out and add
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them together 2 squares for
three squared, 9 four squared 16
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and 5 squared is 25.
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And we can add these very
simply. Foreign 16 is 20.
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9 and 25 is 3420
and 34 is 54.
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What could be simpler?
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Look at these two written using
the Sigma notation, just
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slightly different from the
previous ones. Remember we look
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at all the terms.
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Such as two to the K and we
substitute in the values for K
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in increments of one starting
with zero going up to five.
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So the first one 2 to the K when
K is 0, is 2 to the North. Then
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we add 2 to the one +2 to the
2 + 2 to 3 + 2 to the
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4 + 2 to the five.
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Notice we've got six terms and
the reason why we've got six
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terms is we started K at zero
and not at one.
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And then we evaluate those two
to the zero 1 + 2 plus
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force plus three 2 cubed, which
is it +2 to 416. Let's do
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the 5:32. And when we
add those up, 1 + 2
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is 347-1531.
And then 63.
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Now look at this one.
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A little bit more involved, but
quite straightforward if you
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follow the same steps that we've
done before, we take the general
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term here and we substitute in
values looking at where it
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starts from and where it ends.
So R equals one is the lower
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limit, so we look at our general
term and substituted are equal 1
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first and that will give us one
times by two.
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two. Now we substitute are equal
to for the next one, so that's
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two are plus one is 3 +
1/2. An hour equals 3.
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3 plus One is 4.
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Plus 1/2 an.
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And I forgot my arm, but my are
in this case is 4 and we add 1
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to it which is 5 and continue on
in the same then until we get to
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our upper limit our equals 6. So
that's a half six times by 7.
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And we can evaluate this side,
but we can do a lot of
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canceling on the way before we
do. And if the adding two
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cancers with the two here and
here, 2 cancers with the four
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to give us two here 2 cancers
with this forward to give us 2
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two counters with this six to
give us three and two counters
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with this six to give us 3.
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And then we evaluate eyes adding
together terms. So this one is 1
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plus. And we're left with 1
* 1 * 3, which is 3
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+ 1 * 3 * 2, which
is 6 plus.
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2 * 5 which is 10.
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Plus 3 * 5 which is 15 and
the last 121.
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Now if you look at these
numbers, these are quite
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special numbers. There's a
name given to them, they
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either triangular numbers.
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And you should be able to
recognize these the same way as
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you would recognize square
numbers or prime numbers, or
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order even numbers. I'm finally
what would like to do is just
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add these altogether. 1 + 3 + 6
is 10 + 10 is 20.
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20 + 1535 thirty six and then
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So we get to some 56.
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Now, what
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We've got our general term 2 to
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K equal 1 decay equal an if we
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write it out substituting in our
values for K starting with K
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equal 1. Our first time will be
2 to the one plus then case two
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would be 2 to the 2 + 2
to 3 + 2 to the 4th and
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so on. Until we get to
two to the N.
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Now we can't evaluate
this to a finite number,
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because at this stage we
don't know what Aaron is.
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And what if we
have these two expressions?
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But still using the Sigma
notation, but in this case are
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general term involves a negative
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So we'll just take a little bit
of time and carefully work the
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might. Because it's dead easy to
go wrong in these sorts of ones,
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so take a little bit of time and
you will hopefully work them out
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OK using the same steps that
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we've done before. In this
particular example, our general
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term is negative, one raised to
the part of our and R equals 1
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to 4. So the first term will be
negative. One reason the part of
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one. Plus negative one race to
the power of 2 plus negative
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one. Raise the pirate three and
finally negative one raised to
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the power of 4.
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And then all we have to do is
multiply these item, making sure
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that we get our signs correct,
negative one raise the part of
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one is negative, one negative
one raised to the part of two
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that's squared is positive one.
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Negative 1 cubed.
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That will give negative one and
negative one is part of four.
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That'll give positive one.
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So in fact.
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Are some work site to be O
negative 1 + 1 zero plus
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negative 1 + 1 zero?
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Quite a surprise, in fact, in
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fact. Now look at this one.
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Again, take care. Our general
term is negative one over K
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squared and we have to use
values of K from one to three.
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1. That's negative one over 1
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squared plus negative one over 2
squared plus negative one over 3
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squared. And then again, taking
care with the negative signs.
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Square these values out negative
one over one is negative, 1
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squared will give us one plus.
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Negative one over 2 all squared
that would be positive and is
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1/2 * 1/2, which is quarter.
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Then added two negative one over
3 squared. Again, that's going
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to be positive, and it's going
to be one over 3 * 1 over 3,
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which is one of her and I.
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together.
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The common denominators 36.
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And it will be 9 + 4 and
you've got your one there. So my
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final answer is one and 13 over
36. Quite a grotty answer, but.
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We've been able to workout what
seemed to be quite a complicated
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Sigma notation some.
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And we finally got an answer.
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But what if we had along some
and we wanted to write it in
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Sigma Notation? How would we do
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it? Well, why don't we first
look at the two sums that we
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started off with?
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The first one was 1
+ 2 + 3 +
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4 + 5.
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This one is quite an easy one to
start off with because it's the
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sum. What type of terms? Well,
it's a dead easy term because
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they are adding one on each
time. So if we take the general
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term to BK.
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K starts the lower limit one and
ends with the upper limit 5.
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So there's no problem with this
one. This is very
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straightforward. But look at the
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second one. Again, it's not too
bad because we notice and we
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said before that these are
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square numbers. So we
could actually write these
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as one squared +2 squared,
+3 squared, +4 squared, 5
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squared, and six squared.
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Once we've done this, are Sigma
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because we could easily see our
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general term must be K squared.
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one. Takei equals 6.
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But what
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ones? This is
a long some.
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It's got fractions and it also
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has alternating signs.
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Negative. Positive negative
positive negative positive
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silver. Now the trick in this
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case. Is to rewrite all of these
in terms of fractions.
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And then use negative 1 to a
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power. So if we look at this, we
can rewrite this as negative one
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over 1. Plus 1/2.
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Minus one over 3 +
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1/4 plus. And So what to
one over 100?
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Now we have to deal with
the signs alternating.
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And if we think back we did have
signs alternative before.
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And that's to do with negative
numbers. Now we don't want to
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change the value of the
fraction, we just want to change
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the sign. So if I rewrite
this as negative one.
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Times by one over 1.
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Plus Negative one.
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The power of 2.
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One over 2.
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Plus negative 1 to the power of
3 one over 3.
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Plus negative 1 to the power of
4 one over 4.
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Time 2 plus negative one to
the 101 over 100.
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Have we got the same sum?
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One yes, because
look at each term.
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Negative 1 * 1 over one is
negative, one over 1.
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Negative 1 squared will be
positive 1 * 1/2, which will
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give us our plus 1/2.
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Negative 1 cubed will be
negative one times by 1/3 which
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is negative 1/3, so that's where
we get the negative 1/3 negative
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1 to the power of 4.
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Is positive one times by 1/4
gives us our plus one over 4.
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So rewriting our sequence that
we started off with here.
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Into this format.
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That then helps us to write it
in Sigma Notation.
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Because we can easily see that
this is Sigma.
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We know it's negative one.
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To the power and the power
is always the number underneath
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the fraction. So we'll say that
our fraction is one over K
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and we raise negative 1 to the
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power of K.
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And to complete our Sigma
notation for this long some.
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K equal 1.
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And the upper limit K equal 100.
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So you can easily see that this
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notation. Is very concise.
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And I really lovely way of
writing that big long some.
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Now I'd like to move
on some special cases using
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Sigma notation. The first
special case is when you have to
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sum a constant. Like this?
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If we have Sigma of three
from K equal 1 to 5, what
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does that actually mean?
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Well, we work at night.
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We take each term for cake with
one up to K equal 5.
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Each time remains the same. The
matter whether cake was one K
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equals 2 cakes, 3, four, or
five. So we actually just get
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the sum of 5 threes.
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Which is 15th.
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So with five terms, five times
the constant to make 15.
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So we can get a general result
from that if instead of three we
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have a constant C.
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And we take our
K from one to N.
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What does this work? I to be?
Well, we do it in the same way.
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We workout are some using cake
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with one. 2, three, and so on up
to K equal an.
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And evaluate it.
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So can equal 1 is say K
Equal 2C and so on.
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Up to. The last see
such that we have and terms.
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Because we've got N lots of see
this, some can be simplified
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dead easily to an times C.
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So this is a general result
which is very useful when using
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Sigma notation. If you have to
sum a constant.
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From cable one to N, then the
value is and it is a finite
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value and times by C.
• 21:29 - 21:31
What could be easier?
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• 21:36 - 21:43
What if you had to sum
a variable like this 3K?
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Can this be written more easily?
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Well, we'll take this particular
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example. 3K some 4K. Will
want to four.
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So will sub in cake. Will 1
first, then K equal 2 then K
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equals 3 and then K equals 4 and
we get that some.
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We notice that three is a factor
if each one of these terms.
• 22:13 - 22:20
So we factorize IR 3 and we
get three bracket 1 + 2 +
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3 + 4.
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Now that is equal to three times
by. If you want to add these
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altogether, that's three 610. So
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Now I'm not so much interested
in the answer, but in the
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process. At anyone stage, can
you spot something that we know
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second line. You've got 3 times
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by 1 + 2 + 3 + 4.
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That's the addition of K&K is
one 2, three and four, and we
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can rewrite this.
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Using Sigma notation that
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is 3.
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Sigma. Values
K from K equals
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124.
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So our expression that we
started off with Sigma of three
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K equals 1 to 4K was one
to four equals 3 times.
• 23:30 - 23:36
Sigma of K from cable one to
four. So effectively we've taken
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out with factorized at three
from our first expression, and
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we brought it outside the Sigma
• 23:44 - 23:48
notation. So more generally.
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We can take that if we've got
the Sigma of CK from K equals
• 23:57 - 23:58
1 to N.
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What would that equal?
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Wow, quite easy. All we have to
do is that we see that that is.
• 24:08 - 24:15
See times one plus C times
2 plus C times 3.
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Right up to.
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See times an. We've got M terms
• 24:21 - 24:24
here. We can factorize by the
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sea. And we get 1
+ 2 + 3 plus up
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to N and then we can
rewrite this using Sigma
• 24:37 - 24:44
notation, that is C Sigma
of care from K equals 1
• 24:44 - 24:45
to N.
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And as I say.
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Quite straightforward, when
you've got a Sigma of C times a
• 24:53 - 24:58
variable. See being a constant,
you can take the constant
• 24:58 - 25:03
outside the Sigma notation and
you left with C times by the
• 25:03 - 25:07
Sigma of K from K equal 1 to
• 25:07 - 25:12
N. That's another very useful
result that you should remember
• 25:12 - 25:13
when using Sigma notation.
• 25:14 - 25:20
But what if we had this
expression? The sum of the
• 25:20 - 25:27
variable K plus two where K
takes the lower limit one to
• 25:27 - 25:34
four? What does that work? I'd
be? Can we write it more concert
• 25:34 - 25:36
concisely and easily well?
• 25:37 - 25:41
We do it stage by stage, step by
step, putting in values for K
• 25:41 - 25:43
from K equal 1 to 4.
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So the first time case one says
1 + 2.
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Plus second, her K equals 2, so
that's 2 + 2.
• 25:54 - 26:01
The third term is 3 + 2 and
the fourth term is 4 + 2.
• 26:03 - 26:08
Now what I want to do next is
rearrange these values.
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If I rearrange such that I have
the addition of 1 + 2 + 3 + 4
• 26:16 - 26:23
first. And then see what
we've got left. We've got
• 26:23 - 26:30
that is 4 * 2.
• 26:31 - 26:38
And then we can easily evaluate
this expression 1 + 2 + 3 + 4
• 26:38 - 26:41
is 10 + 4 two 38. So the answer
• 26:41 - 26:47
is 18. But as I said before,
we're not so much concerned with
• 26:47 - 26:52
the answer, but in the process,
can we look and see at any line?
• 26:53 - 26:55
Can we use the Sigma notation
• 26:55 - 26:58
more easily? Well, if we look at
• 26:58 - 27:04
this line. We've seen part
of this before.
• 27:04 - 27:08
1 + 2 + 3 + 4.
• 27:09 - 27:14
Can be written more easily using
Sigma notation as Sigma. Care
• 27:14 - 27:17
from K equals 1 to 4.
• 27:18 - 27:23
Plus then we've got 4
* 2.
• 27:25 - 27:31
So looking back at what we
started with, we've got Sigma of
• 27:31 - 27:38
K plus two from K equals one to
four and it expanded I to this
• 27:38 - 27:45
Sigma notation Sigma of can at
all, from cable one to 4 + 4
• 27:45 - 27:51
* 2 or two is the constant that
• 27:52 - 27:55
So we can generalize
this result.
• 27:57 - 28:01
of.
• 28:02 - 28:09
G of K function of K
plus a constant C.
• 28:09 - 28:17
As our variable and we want to
take care from one to N.
• 28:18 - 28:23
We can use the pattern that
we spotted here to write this
• 28:23 - 28:27
out using Sigma notation plus
some other value.
• 28:28 - 28:32
The first part is Sigma.
• 28:33 - 28:39
Off the function that we started
off with in our variable and the
• 28:39 - 28:40
function is GF care.
• 28:41 - 28:48
And we see that it should be
from K to the upper limit from K
• 28:48 - 28:55
equal 1. Takei equal the upper
limit in this case is for, but
• 28:55 - 28:57
in our general case it's K equal
• 28:57 - 29:01
N. Says K equal 1 to an.
• 29:01 - 29:08
Plus Now the next part, remember
it was the upper limit by the
• 29:08 - 29:14
number of terms times by the
constant that you've had in your
• 29:14 - 29:16
variable. Well, look here. Are
• 29:16 - 29:23
constantly see. And our
upper variable is an.
• 29:23 - 29:26
So it's End Times C.
• 29:27 - 29:30
So this Sigma notation can
• 29:30 - 29:38
be expanded. Two Sigma GF
K from K Equal 1M Plus M
• 29:38 - 29:39
Times C.
• 29:40 - 29:45
And we can even generalize
further using previous results
• 29:45 - 29:50
if we took Sigma from K equals
1 to N.
• 29:50 - 29:54
Of a GF K
• 29:54 - 29:58
Plus C. As our
• 29:58 - 30:01
variable. We can expand that
• 30:01 - 30:08
out. Looking at this time 1st
and then dealing with the
• 30:08 - 30:13
constant now, this term is a
multiple a constant times by
• 30:13 - 30:19
your GF K so we can bring that
outside the Sigma notation.
• 30:19 - 30:27
And say it's a Times by
Sigma GfK from K equals 1
• 30:27 - 30:28
to N.
• 30:30 - 30:31
Plus
• 30:33 - 30:38
Just what's left to do with the
sea and we know when that's
• 30:38 - 30:41
inside the bracket when we
expand it out.
• 30:42 - 30:44
We're left with end times C.
• 30:46 - 30:47
What could be easier?
• 30:48 - 30:51
All right? But
• 30:51 - 30:58
what if? We wanted to
use Sigma notation and the
• 30:58 - 31:03
together is K Plus K squared IE
• 31:03 - 31:05
two functions of K.
• 31:07 - 31:11
What would we get when we write
out this long some?
• 31:12 - 31:14
Start simply by substituting in
• 31:14 - 31:21
for K. Case one to begin with.
So the first term is 1 + 1
• 31:21 - 31:28
squared. Second time is 2 +
2 squared. Third time is 3 + 3
• 31:28 - 31:33
squared, and that's our final
term because the upper limit for
• 31:33 - 31:35
K is 3.
• 31:35 - 31:42
Now we'll rearrange the way we
did before to make our work
• 31:42 - 31:47
easier. Will take the 1 + 2
+ 3 together.
• 31:49 - 31:53
And then we'll take the one
squared and the two squared
• 31:53 - 31:57
and the three squared bits
that are left.
• 31:58 - 32:03
And you can see why I've done
• 32:03 - 32:05
our constant case and here we're
• 32:05 - 32:08
• 32:09 - 32:14
When we add these together, we
just add them up and multiply
• 32:14 - 32:21
out using the square sign you
get 1 + 2 + 3 is 6,
• 32:21 - 32:27
one squared is 1, two squared is
4 three squared 9. Add back
• 32:27 - 32:31
together that gives you
fourteen. 6 + 14 is 20.
• 32:33 - 32:36
But look at this second line.
• 32:37 - 32:43
We can simplify this second line
using Sigma notation.
• 32:44 - 32:49
And it's using Sigma notation
that we've done before. 1 + 2 +
• 32:49 - 32:54
3 is the same as Sigma of care
from K equals 1 to 3.
• 32:55 - 32:56
Plus
• 32:57 - 33:03
1 squared, +2 squared, +3
• 33:03 - 33:09
the sum of all variables, K
squared from K equals 1 to 3.
• 33:10 - 33:14
Now look at what would started
and look what we've got.
• 33:15 - 33:22
We have broken this Sigma
notation up into two sums to
• 33:22 - 33:29
Sigma sums. So here we've
got the Sigma offer variable
• 33:29 - 33:30
plus another variable.
• 33:31 - 33:34
And with split it up to.
• 33:35 - 33:39
Sigma of one variable added to
the Sigma of the other variable.
• 33:40 - 33:42
We've used the distributive law.
• 33:43 - 33:47
So to write this more
• 33:47 - 33:52
generally. We can say that the
sum of a function of care.
• 33:53 - 33:57
function of K.
• 33:59 - 34:05
From K equals 1 to N must
equal the sum.
• 34:06 - 34:13
Two separate sums. The sum of
JFK from K equals 1 to M
• 34:13 - 34:20
Plus Sigma. FFK from K
• 34:22 - 34:27
And in fact, we can continue
this on if we added on other
• 34:27 - 34:28
functions within our variable
• 34:28 - 34:35
here. Then all we have to do
is add on another Sigma of
• 34:35 - 34:36
that particular function.
• 34:38 - 34:43
Finally, here's a real life
example using Sigma notation.
• 34:46 - 34:50
Take for example if we want
to find the mean of a set of
• 34:50 - 34:53
numbers. That could be the
marks in a test.
• 34:54 - 34:59
If we want to find the main what
we want to do is we have to
• 34:59 - 35:00
workout the total sum.
• 35:02 - 35:05
Divided by the number of values.
• 35:06 - 35:11
And that's what the main is.
Well, that's actually what we've
• 35:11 - 35:16
been doing in the previous
examples. We have been finding
• 35:16 - 35:22
the total sum, and we have been
looking at the number of values.
• 35:22 - 35:28
So if I take this example, just
say we had the marks 23456.
• 35:29 - 35:34
And we want to find the mean of
those marks. The main will
• 35:34 - 35:42
equal. 2 + 3 + 4 +
5 + 6 and we'd have to divide
• 35:42 - 35:48
it by the number of values,
which is 5 when we work that
• 35:48 - 35:51
out. That is 5 + 510, another
• 35:51 - 35:57
10. 20 over 5 which works out to
be the value 4.
• 35:58 - 36:00
But more generally.
• 36:00 - 36:07
If you've got a set of marks,
say XI, we can write the main
• 36:07 - 36:14
in terms of Sigma Notation. It's
the total sum of all the marks.
• 36:14 - 36:16
Such as Zhao.
• 36:18 - 36:20
From I equals 1 to N.
• 36:21 - 36:26
And then we want to divide it by
the number of values. Now we
• 36:26 - 36:28
know there are N Marks.
• 36:29 - 36:36
So we want to pie by one over
N, so the main is equal to one
• 36:36 - 36:41
over an Sigma X of I from I
equal 1 to M.
Title:
www.mathcentre.ac.uk/.../Sigma%20notation.mp4
Video Language:
English
 mathcentre edited English subtitles for www.mathcentre.ac.uk/.../Sigma%20notation.mp4