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Error propagation - partial derivative?

  1. Mar 23, 2015 #1
    I am getting a little confused on which error propagation to use:

    I am looking to calculate the error in B*Cos(θ) , for the vertical axis of a williamson hall plot. where B is fwhm of a peak with it's own error and cos of the bragg angle

    I am unsure of whether i need to use partial derivative error propagation

    let Z= B*cos(θ)

    then ΔZ =[ ((dZ/dB)^2 *(ΔB)^2 ) +((dZ/dcosθ)^2 (Δcosθ)^2) ]^0.5

    or standard way

    Δz = [ (ΔB/B)^2 +(Δcosθ/cosθ)^2 ]0.5 * Z

    Any clarification on this would be much appreciated,
    http://file://localhost/Users/isabellasharpley/Library/Caches/TemporaryItems/msoclip/0/clip_image001.gif [Broken]
    Last edited by a moderator: May 7, 2017
  2. jcsd
  3. Mar 23, 2015 #2


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    Staff: Mentor

    Did you measure the error in θ or the error in cos θ?

    If you have the error in cos θ, then either of your two formulas should work. They should give the same result. In fact, you can derive the second formula from the first one.

    If you have the error in θ instead, then you need a formula similar to your first formula, but using dZ/dθ and Δθ instead of dZ/d(cos θ) and Δ(cos θ).
  4. Mar 24, 2015 #3
    Hi jtbell,

    thanks for your response.

    I have an error in θ and also worked out the error in cosθ as Δcosθ = sinθ Δθ.

    the thing is partial derivative formulas (which give the same result) gives me a much larger error compared to the other one, so i'm inclined to go with that one...
  5. Mar 24, 2015 #4
    If the error was measured in, or converted to Cosθ you should not need to differentiate to – Sinθ as the value of Cosθ is just treated as a constant.
  6. Mar 24, 2015 #5
    Z\quad =\quad B\quad \times \quad Cos\theta \\ \frac { \delta Z }{ \delta B } \quad =\quad Cos\theta \\ \frac { \delta Z }{ \delta (Cos\theta ) } \quad =\quad B\\ \\ \Delta Z\quad =\quad (B\quad \times \quad \Delta Cos\theta )\quad +\quad (Cos\theta \quad \times \quad \Delta B)\\

  7. Mar 24, 2015 #6


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    Staff: Mentor


    Then you're calculating one of them wrong. They should give the same result if you calculate them properly. (I made up an example for myself and verified this.) Show your work and someone can probably tell you where you went wrong.
  8. Mar 24, 2015 #7
    thanks Tom_k - So when I compute cos of theta (the bragg angle which has an associated error) the error in it is just the error in theta?

    Even When I put that value in the ordinary error propagation equation, it does not come out equal to the errors in the partial derivatives:

    z= B*Cosθ

    then error via

    Δz= Z* [(ΔB/B)2+(Δcosθ/cosθ)2]1/2

    gives a different result to

    ΔZ = [ ((δZ/δθ)^2 (Δθ)^2) + ((cosθ)^2 (ΔB)^2) ]1/2 = [(Bsinθ)2 (Δθ)2 + (cosθ)2 (ΔB)2 ]1/2

    but either can be used for the error so im going to use the first equation here...
  9. Mar 24, 2015 #8


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    Staff: Mentor

    You haven't shown a sample calculation which demonstrates how the two formulas give different results, but I'll make a wild-assed crazy guess as to why they're different. Are you using degrees instead of radians for θ?
    Last edited: Mar 25, 2015
  10. Mar 25, 2015 #9
    You must get exactly the same result using both methods.
    I will work out a simple example for you.
    Say measurement errors cause B to be measured as 19 mm instead of 20 mm
    And Cosθ is measured as 0.52 rad (29.7738 deg) instead of 0.5236 rad (30 deg)
    ΔB = 1mm = 0.05B ΔCos0 = 0.0036 rad = .0069 Cosθ
    Using partial derivatives:

    [itex]\\ \frac { \delta Z }{ \delta B } \quad =\quad Cos\theta \\ \frac { \delta Z }{ \delta (Cos\theta ) } \quad =\quad B\\ \\ \Delta Z\quad =\quad (B\quad \times \quad \Delta Cos\theta )\quad +\quad (Cos\theta \quad \times \quad \Delta B)\\ \Delta Z\quad =\quad \sqrt { (B\times 0.0069Cos\theta )\^ 2\quad +\quad (Cos\vartheta \times 0.05B)\^ 2 } \\ \Delta Z\quad =\quad 0.0505Z\\[/itex]

    Using other method:

    [itex]\Delta Z\quad =\sqrt { (\frac { \Delta B }{ B } )\^ 2\quad +\quad (\frac { \Delta Cos\vartheta }{ Cos\vartheta } )\^ 2 } \times \quad Z\\ \Delta Z\quad =\quad \sqrt { (\frac { 1 }{ 20 } )\^ 2\quad +\quad (\frac { .0036 }{ .5236 } )\^ 2 } \quad \times \quad Z\\ \Delta Z\quad =\quad 0.0505Z\\[/itex]
  11. Mar 25, 2015 #10
    Yes I had the θ error in degrees JTBell! oops

    Thankyou Tom K, I finally got the errors from both formulas matching!
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