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Integration for Fourier coefficient

  1. May 18, 2015 #1
    1. The problem statement, all variables and given/known data

    I am trying to work out the Fourier coefficient [itex]a_{n}[/itex] for :

    upload_2015-5-18_20-15-36.png

    Mathematics is not my strong point and I would appreciate some help. The answer that wolfram spits out it lovely and neat and I am struggling to get my answer to it.

    2. Relevant equations

    upload_2015-5-18_20-17-46.png

    3. The attempt at a solution
    Apologies for the messy scan, its the best I can do without typing out ALL my working out in LaTeX.
    I am really trying to improve my mathematics, so any suggestions on making my expressions neater, or techniques I may have missed, please do let me know.
    Untitled-1.jpg

    EDIT : it seems uploading has resized my scan.
    heres a https://drive.google.com/file/d/0B9RrInHrjRJCbTJBS0RqYlJHSW8/view?usp=sharing [Broken]

    Thanks for your time
     

    Attached Files:

    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. May 18, 2015 #2

    BvU

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    Hello EL,

    What Fourier series does Wolfram find ? If I fill in n=1 in the expression in the link I don't think I get a useful coefficient.
    It may be lovely and neat, but is it the correct answer for what you are after ?

    You are looking for ## a_n##, using
    upload_2015-5-18_20-17-46-png.83661.png
    for a function that is defined from ##0## to ##2\pi##. (Which is also the range you ask Wolfram to integrate over).

    But you also post

    upload_2015-5-18_20-17-20-png.83659.png
    Does this mean you already know ##b_n=0## or will the sine series come later ?
     
  4. May 19, 2015 #3
    Hi BvU,

    The wolfram link I gave is just for the [itex]a_{n}[/itex]. It is indeterminate at [itex]a_{1}[/itex] and zero for all even numbers thereon. This is fine and to be expected. I have not got round to starting [itex]b_{n}[/itex], I would like to get [itex]a_{n}[/itex] to a workable expression( of [itex]n[/itex]) first. At this moment I am not concerned with the fourier series, just the integral mathematics in working out [itex]a_{n}[/itex]. Have you had a chance to look at my working out for [itex]a_{n}[/itex], does it seem like its on the right track?

    Thanks again
     
  5. May 19, 2015 #4

    BvU

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    Hello again,

    I find it hard to believe a Fourier coefficient can be indeterminate, so I don't think that the elegant result you got from Wolfram is what you where looking for. Check their function "Fourier cos series" (instead of int, and you also remove the *cos etc at the end).

    I can't read your jpg and If I could, then I still can't point to places where I would have questions or remarks. How about if you TeX the first step(s) of what you are trying to do, so we can go over things like:
    what you are trying to integrate
    continuation of the function outside [0,2pi]
    limits
    a few simple cases (n = 1, 2, ...)
    generalization
     
  6. May 19, 2015 #5
    This is the example I am following about the a1 being indeterminate....
    upload_2015-5-19_11-30-44.png
    upload_2015-5-19_11-31-25.png


    Here is the first leg of the calculations :


    [itex]\displaystyle a_{n} = \frac{V}{\pi}\int_{\frac{\pi}{2}}^{\pi}\sin(\omega t)\cos(n\omega t) \delta\omega t + \frac{V}{\pi}\int_{\frac{3\pi}{2}}^{2\pi}\sin(\omega t)\cos(n\omega t)\delta\omega t[/itex]

    So in the first step I have evaluated that between 0 and pi/2 the function equals zero and is omitted, same as with pi to 3pi/2. V has been brought outside the integral.

    [itex]\displaystyle \frac{V}{2\pi}\int_{\frac{\pi}{2}}^{\pi}\sin(\omega t+n\omega t)+\sin(\omega t -n\omega t)\delta\omega t + \frac{V}{2\pi}\int_{\frac{3\pi}{2}}^{2\pi}\sin(\omega t+n\omega t)+\sin(\omega t -n\omega t)\delta\omega t[/itex]

    In step two I have used the "product to sum" identity
    [itex]\displaystyle \sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)][/itex]

    [itex]\displaystyle u=\omega t \pm n\omega t \\ \delta u = n \pm 1 \\ \delta\omega t = \frac{\delta u}{n\pm 1}\\ \\ \frac{V}{2\pi}\Big[-\frac{\cos(\omega t + n \omega t)}{n+1} - \frac{\cos(\omega t - n \omega t)}{n-1} \Big]_{\frac{\pi}{2}}^{\pi} + \frac{V}{2\pi}\Big[-\frac{\cos(\omega t + n \omega t)}{n+1} - \frac{\cos(\omega t - n \omega t)}{n-1}\Big]_{\frac{3\pi}{2}}^{2\pi} \\[/itex]

    Step 3 I have integrated by parts

    [itex]\displaystyle \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(\pi + n \pi)}{n+1} - \frac{\cos(\pi - n \pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{\pi}{2} + n \frac{\pi}{2})}{n+1} - \frac{\cos(\frac{\pi}{2} - n \frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(2\pi + n 2\pi)}{n+1} - \frac{\cos(2\pi - n 2\pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{3\pi}{2} + n \frac{3\pi}{2})}{n+1} - \frac{\cos(\frac{3\pi}{2} - n \frac{3\pi}{2})}{n-1} \Big]\Bigg][/itex]

    step 4 expanding out the limits

    [itex] \displaystyle \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(\pi)\cos(n\pi)- \sin(\pi)\sin(n\pi)}{n+1} - \frac{\cos(\pi)\cos(n\pi)+ \sin(\pi)\sin(n\pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{\pi}{2})\cos(n\frac{\pi}{2})- \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n+1} - \frac{\cos(\frac{\pi}{2})\cos(n\frac{\pi}{2})+ \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(2\pi)\cos(n2\pi)- \sin(2\pi)\sin(n2\pi)}{n+1} - \frac{\cos(2\pi)\cos(n2\pi)+ \sin(2\pi)\sin(n2\pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{3\pi}{2})\cos(n\frac{3\pi}{2})- \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n+1} - \frac{\cos(\frac{3\pi}{2})\cos(n\frac{3\pi}{2})+ \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n-1} \Big]\Bigg][/itex]

    step 5 i have used the sum and difference formula:
    [itex]\displaystyle\cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta[/itex]

    [itex] \displaystyle \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(\pi)\cos(n\pi)}{n+1} - \frac{\cos(\pi)\cos(n\pi)}{n-1} \Big] - \Big[-\frac{- \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n+1} - \frac{ \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(2\pi)\cos(n2\pi)}{n+1} - \frac{\cos(2\pi)\cos(n2\pi)}{n-1} \Big] - \Big[-\frac{- \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n+1} - \frac{ \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n-1} \Big]\Bigg][/itex]

    step 6 I have cancelled out the zero cosine and sine terms

    [itex] \displaystyle \frac{V}{2\pi}\Bigg[\Big[\frac{\cos(n\pi)}{n+1} + \frac{\cos(n\pi)}{n-1} \Big] - \Big[\frac{\sin(n\frac{\pi}{2})}{n+1} + \frac{ \sin(n\frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(n2\pi)}{n+1} - \frac{\cos(n2\pi)}{n-1} \Big] - \Big[-\frac{ \sin(n\frac{3\pi}{2})}{n+1} + \frac{ \sin(n\frac{3\pi}{2})}{n-1} \Big]\Bigg][/itex]

    Step 7 cancelled out the 1 and -1 terms

    [itex] \displaystyle \frac{V}{2\pi}\Bigg[\frac{2n\cos(n\pi)-2n\sin(n\frac{\pi}{2})-2n\cos(n2\pi)+2\sin(n\frac{3\pi}{2}))}{n^2-1}\Bigg][/itex]

    step 8 consolidated

    Here is where I am stuck to carry on....
    inputting my last expression into a wolfram table reveals ....LINK
    In comparison to LINK which is an answer given in a previous topic on PF .....LINK

    Somewhere I have made a mistake!

    Thanks again
    EL
     
  7. May 19, 2015 #6

    SammyS

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    You left out a factor of n from the coefficient of the last sine term in the numerator.

    It should be ## \ 2n\sin(n\frac{3\pi}{2}) \ ##.

    Added in Edit:

    To find a_1, simply put n = 1 in the integral expression. Then evaluate that.

    Link to Wolfram inserting n = 1 in your original expression.
     
    Last edited: May 19, 2015
  8. May 19, 2015 #7
    Hi SammyS,

    I got the [itex]- \frac{2\, \sin\!\left(\frac{3\, n\, \mathrm{pi}}{2}\right)}{\left(n^2 - 1\right)}[/itex] from simplifying the previous part of the expression [itex]\frac{\sin\!\left(\frac{3\, n\, \mathrm{pi}}{2}\right)}{\left(n + 1\right)} - \frac{\sin\!\left(\frac{3\, n\, \mathrm{pi}}{2}\right)}{\left(n - 1\right)}[/itex]
    Simplifying gives :
    [itex]- \frac{2\, \sin\!\left(\frac{3\, n\, \mathrm{pi}}{2}\right)}{\left(n^2 - 1\right)}[/itex]

    added back into the expression and corrected ...
    [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\frac{2n\cos(n\pi)-2n\sin(n\frac{\pi}{2})-2n\cos(n2\pi)-2\sin(n\frac{3\pi}{2}))}{n^2-1}\Bigg]
    [/itex]
    Sadly this does not yield the results I am looking for, but it seems closer.
    I see I have missed the last minus sign, but I don't see how it can equal [itex] 2n\sin(n\frac{3\pi}{2})[/itex].
    Do you mean to say that I have made a mistake previous to this?

    thanks
     
  9. May 19, 2015 #8

    SammyS

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    I think I was mainly looking at patterns, but now that you mention it: There is an error going from step 5 to step 6..

    In eqach of the two groups of expressions with sin() , the sines have opposite sign. That is what cancels n & gives a two.

    So in step 6, that first sin() expression should not have a coefficient n, just 2 .
     
  10. May 19, 2015 #9
    Sorry I am not following you, can you show me more visually what you mean?
     
  11. May 19, 2015 #10
    1. The problem statement, all variables and given/known data
    I am trying to work out the Fourier coefficient [itex]a_{n}[/itex]for :

    upload_2015-5-18_20-15-36-png.83658.png

    This question has been asked in a previous thread HERE

    2. Relevant equations
    upload_2015-5-18_20-17-46-png.83661.png

    3. The attempt at a solution

    [itex]
    \displaystyle a_{n} = \frac{V}{\pi}\int_{\frac{\pi}{2}}^{\pi}\sin(\omega t)\cos(n\omega t) \delta\omega t + \frac{V}{\pi}\int_{\frac{3\pi}{2}}^{2\pi}\sin(\omega t)\cos(n\omega t)\delta\omega t
    [/itex]

    So in the first step I have evaluated that between 0 and pi/2 the function equals zero and is omitted, same as with pi to 3pi/2. V has been brought outside the integral.

    [itex]
    \displaystyle \frac{V}{2\pi}\int_{\frac{\pi}{2}}^{\pi}\sin(\omega t+n\omega t)+\sin(\omega t -n\omega t)\delta\omega t + \frac{V}{2\pi}\int_{\frac{3\pi}{2}}^{2\pi}\sin(\omega t+n\omega t)+\sin(\omega t -n\omega t)\delta\omega t
    [/itex]

    In step two I have used the "product to sum" identity
    [itex]\displaystyle \sin\alpha\cos\beta = \frac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)][/itex]


    [itex]
    \displaystyle u=\omega t \pm n\omega t \\ \delta u = n \pm 1 \\ \delta\omega t = \frac{\delta u}{n\pm 1}\\ \\ \frac{V}{2\pi}\Big[-\frac{\cos(\omega t + n \omega t)}{n+1} - \frac{\cos(\omega t - n \omega t)}{n-1} \Big]_{\frac{\pi}{2}}^{\pi} + \frac{V}{2\pi}\Big[-\frac{\cos(\omega t + n \omega t)}{n+1} - \frac{\cos(\omega t - n \omega t)}{n-1}\Big]_{\frac{3\pi}{2}}^{2\pi} \\
    [/itex]

    Step 3 I have integrated by parts

    [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(\pi + n \pi)}{n+1} - \frac{\cos(\pi - n \pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{\pi}{2} + n \frac{\pi}{2})}{n+1} - \frac{\cos(\frac{\pi}{2} - n \frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(2\pi + n 2\pi)}{n+1} - \frac{\cos(2\pi - n 2\pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{3\pi}{2} + n \frac{3\pi}{2})}{n+1} - \frac{\cos(\frac{3\pi}{2} - n \frac{3\pi}{2})}{n-1} \Big]\Bigg]
    [/itex]

    step 4 expanding out the limits

    [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(\pi)\cos(n\pi)- \sin(\pi)\sin(n\pi)}{n+1} - \frac{\cos(\pi)\cos(n\pi)+ \sin(\pi)\sin(n\pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{\pi}{2})\cos(n\frac{\pi}{2})- \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n+1} - \frac{\cos(\frac{\pi}{2})\cos(n\frac{\pi}{2})+ \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(2\pi)\cos(n2\pi)- \sin(2\pi)\sin(n2\pi)}{n+1} - \frac{\cos(2\pi)\cos(n2\pi)+ \sin(2\pi)\sin(n2\pi)}{n-1} \Big] - \Big[-\frac{\cos(\frac{3\pi}{2})\cos(n\frac{3\pi}{2})- \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n+1} - \frac{\cos(\frac{3\pi}{2})\cos(n\frac{3\pi}{2})+ \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n-1} \Big]\Bigg]
    [/itex]

    step 5 i have used the sum and difference formula:
    [itex]
    \displaystyle\cos(\alpha\pm\beta) = \cos\alpha\cos\beta \mp \sin\alpha\sin\beta
    [/itex]

    [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(\pi)\cos(n\pi)}{n+1} - \frac{\cos(\pi)\cos(n\pi)}{n-1} \Big] - \Big[-\frac{- \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n+1} - \frac{ \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(2\pi)\cos(n2\pi)}{n+1} - \frac{\cos(2\pi)\cos(n2\pi)}{n-1} \Big] - \Big[-\frac{- \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n+1} - \frac{ \sin(\frac{3\pi}{2})\sin(n\frac{3\pi}{2})}{n-1} \Big]\Bigg]
    [/itex]

    step 6 I have cancelled out the zero cosine and sine terms

    [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\Big[\frac{\cos(n\pi)}{n+1} + \frac{\cos(n\pi)}{n-1} \Big] - \Big[\frac{\sin(n\frac{\pi}{2})}{n+1} + \frac{ \sin(n\frac{\pi}{2})}{n-1} \Big]\Bigg] + \frac{V}{2\pi}\Bigg[\Big[-\frac{\cos(n2\pi)}{n+1} - \frac{\cos(n2\pi)}{n-1} \Big] - \Big[-\frac{ \sin(n\frac{3\pi}{2})}{n+1} + \frac{ \sin(n\frac{3\pi}{2})}{n-1} \Big]\Bigg]
    [/itex]

    Step 7 cancelled out the 1 and -1 terms

    [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\frac{2n\cos(n\pi)-2n\sin(n\frac{\pi}{2})-2n\cos(n2\pi)-2\sin(n\frac{3\pi}{2}))}{n^2-1}\Bigg]
    [/itex]

    step 8 consolidated

    Here is where I am stuck to carry on.... I I have made a mistake somewhere because inputting my expression doesn't yield the same results as the answer given in the old thread.
    Maths isn't my strong point and I would love to know where I have gone wrong.

    Thanks again
    EL
     
  12. May 19, 2015 #11

    berkeman

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    (2 threads merged)
     
  13. May 19, 2015 #12

    SammyS

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    The part of step 5 I was referring to is:
    ##\displaystyle \ - \left[-\frac{- \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n+1} - \frac{ \sin(\frac{\pi}{2})\sin(n\frac{\pi}{2})}{n-1} \right] \ ##

    What you should have for the corresponding part of step 6 is:
    ##\displaystyle \ - \left[\frac{\sin(n\frac{\pi}{2})}{n+1} - \frac{ \sin(n\frac{\pi}{2})}{n-1} \right] \ ##

    not
    ##\displaystyle \ - \left[\frac{\sin(n\frac{\pi}{2})}{n+1} + \frac{ \sin(n\frac{\pi}{2})}{n-1} \right] \ ##
     
  14. May 20, 2015 #13
    Thanks for the continual effort SammyS!

    The corrected final term is [itex]
    \displaystyle \frac{V}{2\pi}\Bigg[\frac{2n\cos(n\pi)+2\sin(n\frac{\pi}{2})-2n\cos(n2\pi)-2\sin(n\frac{3\pi}{2}))}{n^2-1}\Bigg]
    [/itex]

    Matlab simplifys this to [itex]\frac{2\, v\, \sin\!\left(\frac{n\, \mathrm{pi}}{2}\right)\, \left(2\, {\sin\!\left(\frac{n\, \mathrm{pi}}{2}\right)}^2 + n\, \sin\!\left(\frac{3\, n\, \mathrm{pi}}{2}\right) - 1\right)}{\mathrm{pi}\, \left(n^2 - 1\right)}[/itex]

    I'm not sure how this has been simplified, any ideas?

    The new expression yields almost identical results to the compared answer, except my results are all negative.

    my results:
    n=1 gives indeterminate
    n=3 gives -v/pi
    n=5 gives -v/(3*pi)
    n=7 gives -v/(3*pi)
    n=9 gives -v/(5*pi)

    In comparison to the "answer" table
    LINK
    MSP17121f96cife4i2gdcgc00002434g869827d62c7.gif
    Also... a comparison between the actual expression worked out by wolfram LINK
    MSP29121dce93e162eg538100001d1929d5eha1ihi0.gif
    Compared with my final expression simplified in wolfram LINK
    MSP78541i2bbffb083c9h0500003he2be18i7ecac6g.gif
    gives MSP78621i2bbffb083c9h0500004c7905ecgi4h2900.gif

    I can see the differences are that one expression is negative, the other isn't (explaining why I have negative answers) and the top part of the expression, N is multiplied by the cos term, not the sin term.

    So close!

    I will have another thorough check of all my workings out later today, but do you have any insight into how I can simplify the final expression as much as possible, so that it resembles something like what matlab/wolfram spits out?

    Thanks again
     
  15. May 20, 2015 #14

    SammyS

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    I get the impression that you're stumped in two ways. One is the indeterminate nature of coefficient, a1 . The other is that you're not quite producing the correct values for the other coefficients.

    Let's look at what's going on with a1.

    I did suggest in a previous post, that you simply evaluate you're initial integral directly with a value of 1 inserted for n.
    but...

    Let's what it is about n = 1 that causes the general result you have to be in error. (the indeterminate thing)

    It's just the (n-1) factor which produces the problem. So let's consider the following indefinite integral.
    ##\displaystyle\ \int \sin(\omega t -n\omega t)\delta\omega t ##​

    First of all, if n = 1, then the integrand becomes sin(0) which is zero, thus the definite integral resulting from this is zero.

    If you consider the substitution, ##\ u=\omega t - n\omega t\ ##, you get ##\displaystyle\ \delta u = (1-n)\delta\omega t\ ##. (Maybe this is the source of your sign error.)

    When you solve for ##\ \delta\omega t\ ##, you get ##\displaystyle\ \delta\omega t = \frac{\delta u}{1-n}\ ##.

    See the problem if n = 1 ?
     
    Last edited: May 20, 2015
  16. May 20, 2015 #15
    Ok yes, got it, thanks for that! I can see how [itex]a_{1}=-\frac{v}{pi}[/itex]. Sadly I haven't had a time today to check my expression, but I will tomorrow, I'm sure I will find the mistake!

    Thanks again
    EL
     
  17. May 21, 2015 #16
    Ok I have managed to get the correct results, after correcting the substitution to 1-n and rewriting the expression. Thanks for your help!

    I now have simplified the expression as much as I can to
    [itex]\displaystyle\frac{v\bigg(\cos(n\pi)+n\sin(n\frac{\pi}{2})-\cos(n2\pi)-n\sin(n\frac{3\pi}{2})\bigg)}{\pi(1-n^2)}[/itex]

    Yet wolfram still simplifies it even more into :
    [itex] - \frac{2\, v\, \sin\!\left(\frac{n\, \mathrm{\pi}}{2}\right)\, \left(\sin\!\left(\frac{3\, n\, \mathrm{\pi}}{2}\right) - n\, \cos\!\left(n\, \mathrm{\pi}\right)\right)}{\mathrm{\pi}\, \left(n^2 - 1\right)}[/itex]

    Any idea how it achieved this? A trig identity? I couldn't see where any would help me.

    After this I'll be well prepared for working out [itex]b_{n}[/itex]!! :woot:

    Thanks again
    EL
     
  18. May 21, 2015 #17

    SammyS

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    Difference to product.
     
  19. May 21, 2015 #18
    Ah ok got it now. Thanks very much for all the help!

    EL
     
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