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Thread: Summarising an Equation

  1. #1
    Junior Member
    Mar 2009

    Summarising an Equation


    Can someone please explain on how the below equations were summarised?

    Summarising an Equation-equation-1.jpg
    I can see in the above equation that 3 is common to all the elements so that can go outside the bracket Ė the rest catches me a bit though.

    Summarising an Equation-equation-2.jpg

    Same in equation 2 Ė I can see 3/2 and are (2 + square root(2)) are common to every element ..its the rest I donít get.

    Thanks in advance
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  2. #2
    MHF Contributor
    Nov 2008

    Equation 1
    $\displaystyle \displaystyle 3 \cdot 1 + 3 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} + \cdots + 3 \cdot \frac{1}{2^{k-2}} = 3 \cdot \left[\left(\frac{1}{2}\right)^0 + \left(\frac{1}{2}\right)^1 + \cdots + \left(\frac{1}{2}\right)^{k-2}\right]$

    The sum is a sum of consecutive terms of a geometric series

    $\displaystyle \displaystyle 3 \cdot 1 + 3 \cdot \frac{1}{2} + 3 \cdot \frac{1}{2} + \cdots + 3 \cdot \frac{1}{2^{k-2}} = 3 \cdot \frac{1-\left(\frac{1}{2}\right)^{k-1}}{1-\frac{1}{2}}\right] = 3 \cdot \left(2 - \left(\frac{1}{2}\right)^{k-2}\right)$

    Equation 2
    It is the sum of (k-1) terms all equal to $\displaystyle \displaystyle \frac{3}{2} \cdot (2+\sqrt{2})$ therefore it is $\displaystyle \displaystyle \frac{3}{2} \cdot (2+\sqrt{2})(k-1) $
    Last edited by running-gag; Jul 14th 2010 at 07:48 AM. Reason: Precision added
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  3. #3
    Super Member

    May 2006
    Lexington, MA (USA)
    Hello, kerrymaid!

    These problems are formatted badly and have typos.
    And the second one is downright silly.

    $\displaystyle 3\cdot1 + 3\cdot \frac{1}{2} + 3\cdot\frac{1}{2^2} + \hdots + 3\cdot\frac{1}{2^{k-2}} \;=\;3 \left(2 - \frac{1}{2^{k-2}}\right)$

    $\displaystyle \text{The left side is: }\;3\underbrace{\left[1 + \frac{1}{2} + \frac{1}{2^2} + \hdots + \frac{1}{2^{k-1}}\right]}_{\text{geometric series}} $ .[1]

    The series has:
    . . first term $\displaystyle a = 1$, common ratio $\displaystyle r = \frac{1}{2}$, and $\displaystyle n = k\!-\!1$ terms.

    It sum is: .$\displaystyle S \;=\;a\,\frac{1-r^n}{1-r}$

    . . $\displaystyle S \;=\;1\cdot\dfrac{1-(\frac{1}{2})^{k-1}}{1-\frac{1}{2}} \;=\;\dfrac{1 - \frac{1}{2^{k-1}}}{\frac{1}{2}} \;=\;2 - \frac{2}{2^{k-1}} \;=\;2 - \frac{1}{2^{k-2}} $

    Therefore, [1] become: .$\displaystyle 3\left(2 - \frac{1}{2^{k-2}}\right)$

    $\displaystyle \left[(2\!+\!\sqrt{2})\cdot\frac{3}{2}\right] + \left[\frac{1}{2}(2\!+\!\sqrt{2})\cdot2\cdot\frac{3}{2}\ right] + \left[\frac{1}{2^2}(2\!+\!\sqrt{2})\cdot 2^2\cdot\frac{3}{2}\right] + \hdots$

    . . . . $\displaystyle + \left[\frac{1}{2^{k-2}}(2\!+\!\sqrt{2})\cdot2^{k-2}\cdot \frac{3}{2}\right] \;\;=\;\;\frac{3}{2}(2\!+\!\sqrt{2})(k-1)$

    . . This is really silly!

    The second term has: .$\displaystyle \frac{1}{2}\cdot2\:=\:1$

    The third term has: .$\displaystyle \frac{1}{2^2}\cdot2^2\:=\:1$

    The last term has: .$\displaystyle \frac{1}{2^{k-2}}\cdot2^{k-2} \:=\:1 $

    $\displaystyle \text{The sum becomes: }\;\underbrace{\tfrac{3}{2}(2\!+\!\sqrt{2}) + \tfrac{3}{2}(2\!+\!\sqrt{2}) + \hdots + \tfrac{3}{2}(2\!+\!\sqrt{2})}_{k-1\text{ terms}} $

    . . . . . . . . . . . . . $\displaystyle =\;\frac{3}{2}(2\!+\!\sqrt{2})(k-1)$

    Edit: Too slow . . . again!
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  4. #4
    Junior Member
    Mar 2009
    Thanks very much to Soroban and running-gag - I understand now.
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  5. #5
    Junior Member
    Mar 2009

    As an extension to the above question, I'm hoping someone can help me to understand the following (sorry for the long email and my rubbish use of proper notation - havent figured out how to easily do that in this editor yet). Iím not sure if this is strictly a pure algebra question but Iím hoping someone can help just the same.

    Iíve attached a diagram which shows a square divided into a grid hierarchy h=3.
    Four of the smallest squares make up an order 1 square, four order 1 squares make up an order 2 square, four order 2 squares make up an order 3 square and so on.

    The side length of an order 1 square is L which is the transmission range of a device i.e. 200m. A device has a server in each sibling square at each hierarchy level i.e. a device has 3 servers in its order 1 square, 3 in the order 2 square and so on. A device updates its servers in order i sibling squares after movement of 2i-1.d where d is the update threshold.

    Therefore , the number of hops travelled by all update packets per interval ends up being Equation 2 above(Thanks again to soroban and running-gag for helping me to understand this)

    I understand the (2 + sqrt(2)) comes from the number of hops travelled to the 3 servers within an order 1 square (1 + 1 + sqrt(2) for the diagonal sibling square). The 2^0 + 2^1 + 2^2 Ö + 2^k-2 comes from the side length L i.e. 200m of a square which doubles as the hierarchy increases. The 1 + Ĺ + ľ Ö + 1/(2^k-2) comes from the number of updates sent per one interval (As stated above, the frequency of updates gets progressively less for servers in more remote squares). What I canít grasp is where the 3/2 comes from.

    Iíve included below the explanation from the paper Iím reading which I cannot fathom as I would have thought the above equation (without adding an additional factor of 3/2) would adequately cover an equation to approximate the ďnumber of hops travelled by all update packets per intervalĒ.

    The factor of 3/2 for each term is explained as follows. Let the lowest-order sibling squares containing the destination and the server be order-k squares X and Y, respectively. The update packet is first forwarded in a straight line towards Y while inside X. Once crossing the boundary between X and Y, it is forwarded in a sequence of steps in traversing up the hierarchy starting from an order-1 square and reaching the order-(k - 1) square containing the server. Since the sequence of steps are directionless and the expected distances
    they travel are recursively doubled, they incur an approximate factor of 2 overhead compared to the direct distance between the starting point and finishing point inside square Y. The factor of 2 overhead while traversing inside Y and the straight line path inside X together contribute to the factor of 3/2 for each term in the above average hop count formula for update packets.
    If anyone can grasp what is meant by this and why it should be included in the equation I would really appreciate the help.

    Many thanks in advance.
    Attached Thumbnails Attached Thumbnails Summarising an Equation-grid.bmp  
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