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Math Help - calculation of ratio with random error

  1. #1
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    calculation of ratio with random error

    I have trouble solving a problem assigned to us for class and I dont even know how to start. The problem goes as follows: We look at a guitar string and its frets (numerated from fret 0 (the whole string) to fret 12 (half). The string is 600 mm long. Now the frets are arranged such that the ratio from fret n / fret n+1 = 2^(1/12). (The following setup was created by me, it might therefore be wrong.) So the position of fret n (denoted by a_0):

    a_n = 600 \mdot (2^{-\frac{1}{12}})^n

    Here's the catch. Now in an experiment we measure the position of all frets. We will by by that have a random measurement error and get:

    a'_n = a_n + \epsilon_n where \epsilon_n \in [-0.5,0.5]

    Now I am supposed to explain why the average of all the ratios:

    \frac{1}{12}\sum_0^{11}\frac{a'_n}{a'_{n+1}}

    improves my accuracy by a factor 3-4. I get the intition but to state numbers, I tried to solve this filling in what I know but even MATLAB only gives me unreadable results on this. Furthermore I am expected to explain why this only derives from 2^{-1/12} by no more than 0.0000004.

    Any ideas? Thanks a lot, I am lost on this.
    Last edited by raphw; October 9th 2010 at 09:55 PM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by raphw View Post
    I have trouble solving a problem assigned to us for class and I dont even know how to start. The problem goes as follows: We look at a guitar string and its frets (numerated from fret 0 (the whole string) to fret 12 (half). The string is 600 mm long. Now the frets are arranged such that the ratio from fret n / fret n+1 = 2^(1/12). (The following setup was created by me, it might therefore be wrong.) So the position of fret n (denoted by a_0):

    a_n = 600 \mdot (2^{-\frac{1}{12}})^n

    Here's the catch. Now in an experiment we measure the position of all frets. We will by by that have a random measurement error and get:

    a'_n = a_n + \epsilon_n where \epsilon_n \in [-0.5,0.5]

    Now I am supposed to explain why the average of all the ratios:

    \frac{1}{12}\sum_0^{11}\frac{a'_n}{a'_{n+1}}

    improves my accuracy by a factor 3-4. I get the intition but to state numbers, I tried to solve this filling in what I know but even MATLAB only gives me unreadable results on this. Furthermore I am expected to explain why this only derives from 2^{-1/12} by no more than 0.0000004.

    Any ideas? Thanks a lot, I am lost on this.
    Write a'_n=a_n+\varepsilon_n, where \varepsilon_n \sim U(-0.5,0.5) for n=1, .., 12 and \varepsilon_0=0.

    Now expand

    \displaystyle \frac{1}{12}\sum_0^{11}\dfrac{a'_n}{a'_{n+1}}=\dfr  ac{1}{12}\sum_0^{11}\dfrac{a_n+\varepsilon_n}{a_{n  +1}+\varepsilon_{n+1}}

    upto the terms linear in \varepsilon_i,\ i=1,..n.

    Now estimate the error (or rather the standard error of the estimate).

    CB
    Last edited by CaptainBlack; October 11th 2010 at 06:44 AM. Reason: put missing 1/12 back in
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  3. #3
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    Hello CaptainBlack,
    thank you for your answer. However, this is exactly where I have problems. I already expanded my formula an tried to get the Epsilons out. But all the resulting squares make it hard to find a result. I put this whole thing into MATLAB:

    Code:
    clear;clc;
    
    % a : The true values for the frats
    % ar: (1) The true values for the ratios (2) tratios distorted by +0.5 (3)
    % the ratios distorted by -0.5 (4/5) crossed ratios (6/7) alternating
    % ratios
    % ar_maxmin : the extreme values of the ratios
    % avg : the average of the ratios with respect to the calculation done for
    % ar
    % rerror : the deviation of the ratios to the 12th squareroot of 2
    
    a = zeros(3,13);
    for i = 1:13
        a(1,i) = 600*(2^(-1/12))^(i-1);
        a(2,i) = a(1,i) + 0.5;
        a(3,i) = a(1,i) - 0.5;
    end
    
    ar = zeros(6,12);
    
    for i = 1:12
        ar(1,i) = a(2,i)/a(2,i+1);
        ar(2,i) = a(3,i)/a(3,i+1);
        ar(3,i) = a(2,i)/a(3,i+1);
        ar(4,i) = a(3,i)/a(2,i+1);
        ar(5,i) = a(2 + mod(i,2),i)/a(2 + mod(i+1,2),i+1);
        ar(6,i) = a(2 + mod(i+1,2),i)/a(2 + mod(i,2),i+1);
    end
    
    ar_maxmin = [max(ar,[],2) min(ar,[],2) max(ar,[],2)-min(ar,[],2)];
    avg = mean(ar,2)
    rerror = abs([2^(1/12); 2^(1/12); 2^(1/12); 2^(1/12); 2^(1/12); 2^(1/12)] - avg);
    
    clear i;
    what gives me approximatly the same results that i need. However, I see no way of solving that by pure theory. I also tried to approximate the result by getting rid of some of the terms and deriving a higher number to find an upper bound of the error but I solved that the error would not be higher then 6000 mm (surprise...!) what does not really lead me anywhere. Could you therefore be a little more spesific? Im not lazy, promised, I just cannot see what you mean. Thank you!
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack View Post
    Write a'_n=a_n+\varepsilon_n, where \varepsilon_n \sim U(-0.5,0.5) for n=1, .., 12 and \varepsilon_0=0.

    Now expand

    \displaystyle \frac{1}{12}\sum_0^{11}\dfrac{a'_n}{a'_{n+1}}=\dfr  ac{1}{12}\sum_0^{11}\dfrac{a_n+\varepsilon_n}{a_{n  +1}+\varepsilon_{n+1}}

    upto the terms linear in \varepsilon_i,\ i=1,..n.

    Now estimate the error (or rather the standard error of the estimate).

    CB
    Quote Originally Posted by raphw View Post
    Hello CaptainBlack,
    thank you for your answer. However, this is exactly where I have problems. I already expanded my formula an tried to get the Epsilons out. But all the resulting squares make it hard to find a result. I put this whole thing into MATLAB: ....
    \dfrac{a_n+\varepsilon_n}{a_{n+1}+\varepsilon_{n+1  }}=\dfrac{a_n(1+\frac{\varepsilon_n}{a_n})}{a_{n+1  }(1+\frac{\varepsilon_{n+1}}{a_{n+1}})}<br />
\approx \dfrac{a_n}{a_{n+1}}\left( 1+\frac{\varepsilon_n}{a_n} \right)\left(1-\frac{\varepsilon_{n+1}}{a_{n+1}} \right)}<br />

    ............... <br />
= \dfrac{a_n}{a_{n+1}}\left[1+\frac{\varepsilon_n}{a_n}-\frac{\varepsilon_{n+1}}{a_{n+1}} +O(\varepsilon^2) \right]=2^{1/12}\left[1+\frac{\varepsilon_n}{a_n}-\frac{\varepsilon_{n+1}}{a_{n+1}} +O(\varepsilon^2) \right]

    So ignoring second degee terms in \varepsilon we have the original sum is a telescoping sum in the \varepsilon s so:

    \displaystyle \frac{1}{12}\sum_0^{11}\dfrac{a'_n}{a'_{n+1}}=2^{1/12}+\frac{2^{1/12}}{12}\left[\frac{\varepsilon_0}{a_0}-\frac{\varepsilon_{12}}{a_{12}} \right]

    hence this sum has expectation 2^{1/12} and standard deviation:

    \dfrac{2^{1/12}}{12\times \sqrt{12}\times a_{12}}\approx 8.5 \times 10^{-5}

    (here we are using the fact that the SD of a RV uniform on an interval of length 1 is 1/\sqrt{12} , and that \varepsilon_0=0)

    The details of the above you will need to check.

    CB
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  5. #5
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    Thanks a ton!
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