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

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    This video is all about Cigna
    notation. This is a concise and
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    precise way of explaining long
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    sums. We study some examples and
    some special cases and then will
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    drive some general results.
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    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
    lovely answer 100.
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    What about this one? It's
    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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    Times by 1/2 added
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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
    add the 20 is 56.
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    So we get to some 56.
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    Now, what
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    about this?
    We've got our general term 2 to
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    the K and we want to start with
    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
    sign in both cases.
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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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    So we start with K equal
    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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    We can add these together,
    adding the two fractions
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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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    notation some? Is dead easy
    because we could easily see our
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    general term must be K squared.
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    And K must start with cake, will
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    one. Takei equals 6.
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    But what
    about these
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    ones? This is
    a long some.
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    What's different about this?
    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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    because we start with K equal 1
    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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    We start with the lower limit of
    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.
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    What could be easier?
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    But what about this?
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    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.
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    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
    the answer is actually 30.
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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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    already? Well, look at this
    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.
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    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
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    notation. So more generally.
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    We can take that if we've got
    the Sigma of CK from K equals
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    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.
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    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
  • 24:24 - 24:31
    sea. And we get 1
    + 2 + 3 plus up
  • 24:31 - 24:37
    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.
  • 24:46 - 24:48
    And as I say.
  • 24:48 - 24:53
    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.
  • 25:43 - 25:47
    So the first time case one says
    1 + 2.
  • 25:49 - 25:54
    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.
  • 26:09 - 26:16
    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
    add 2 four times, so
    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
    was added to the care.
  • 27:52 - 27:55
    So we can generalize
    this result.
  • 27:57 - 28:01
    If we had Sigma
    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
    variable we wanted to add
    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
    that, because here we're adding
  • 32:03 - 32:05
    our constant case and here we're
  • 32:05 - 32:08
    adding. Our case squareds.
  • 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
    squared, that's dead easy. It's
  • 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
    Added to another
    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
    equals 1 to add.
  • 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
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