Gauss' Approach

Discussion in 'Physics & Math' started by Thoreau, Jun 1, 2011.

  1. Thoreau Valued Senior Member

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    Can someone please explain Gauss' Approach, step by step? I have sat here for the past hour trying to understand it and I just can't seem to wrap my head around it. I've looked at many online resources, none of which sufficiently explain the steps.
     
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  3. rpenner Fully Wired Valued Senior Member

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    The definition of the "mean value" (average value, expected value) for a nonempty, finite, set of numbers is the "sum of the set" divided by the "count of the set".

    \(\textrm{mean value} = \frac{\textrm{sum of the set}}{\textrm{count of the set}}\)

    Gauss turned this into a useful tool when finding out the mean value was easier than finding the sum of the set.

    \(\textrm{sum of the set} = \left( \textrm{count of the set} \right) \times \left( \textrm{mean value} \right)\)

    Let's say the set is the first ten positive integers: \(S = \left{ 1, 2, 3, 4, 5, 6, 7, 8, 9, 10 \right}\). The count is 10, \(\textrm{card}(S) = 10\), and the mean value is 5.5. \(E(S) = \frac{\sum_{k \in S} k }{\textrm{card} (S)}= \frac{1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 }{10} = \frac{55}{10} = 5.5\). As you see, Gauss's Approach is useless unless you can find a smarter way of finding the mean value than doing the addition.

    Gauss's trick works for all unbroken finite sequences where the elements are separated by a constant difference, i.e. for arithmetic sequences. He noted that the mean value of the whole sequence is the same as the mean value of just the first and last element of the sequence, thus \(E(S) = \frac{1 + 10}{2} = \frac{11}{2} = 5.5\). Thus the sum of the first ten positive integers is \(\sum_{k \in S} k = \textrm{sum of the set} = \left( \textrm{count of the set} \right) \times \left( \textrm{mean value} \right) = \left( \textrm{card}(S) \right) \times \left( E(S) \right) = 10 \times \frac{1 + 10}{2} = 5 \times 11 = 55\).

    A bit of gymnastics which seems unnecessary for Gauss, but his teacher assigned him the take of summing the first hundred numbers, thinking to get some peace and quiet. Gauss computed: \(\left( \textrm{card}(S) \right) \times \left( E(S) \right) = 100 \times \frac{1 + 100}{2} = 50 \time 101 = 5050\) and had the rest of the day to play.

    The trick also works for any unbroken arithmetic sequence with minor enhancements. Lets say the sequence is \( S = \left{ a_1, a_2, a_3, \dots a_n \right} = \left{ a_1, a_1 + b, a_1 + 2b, \dots a_1 + (n-1) b \right}\) then \(\textrm{card}(S) = n\) and \(E(S) = a_1 + \frac{n-1}{2} b\), so \(\sum_{k \in S} k = n \times \left( a_1 + \frac{n-1}{2} b \right) = n \times a_1 + \frac{n \times (n-1)}{2} b\) where \(a_1\) is the first element of the sequence and \(b\) is the step between the elements.

    So now you can show that the sum of the first million positive even numbers is exactly 1,000,001,000,000.
     
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  5. Dinosaur Rational Skeptic Valued Senior Member

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    The story I heard was that the teacher asked for the sum of the first 1000 integers & young Gauss thought as follows.
    It is donkey work to add up the 1000 integers. Anyone with the ability to do mathematics hates donkey work & looks for some clever method.

    Young Gauss did some doodling with integers.

    His doodling led to making pairs of integers: 1 & 1000, 2 & 999, 3 & 998, et cetera.

    He noticed that each pair added up to 1001 & there were 500 pairs.

    Ergo, the sum is 500 *1001 = 500500​
    I do not think he was enough of a genius to quickly derive the general formula for the sum of an arithmetic series. He was only 6-8 years old at the time
     
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  7. Thoreau Valued Senior Member

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    Thank you both.
     

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