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Use differentials to estimate the maximum possible error

  1. Oct 24, 2004 #1
    Four positive numbers, each less than 50, are rounded to the first decimal place and then multiplied together. Use differentials to estimate the maximum possible error in the computed product that might result from the rounding.

    I took f(w,x,y,z) = wxzy and then you set df=xzy(dw)+wzy(dx) etc and you will eventually wind up with 4(50)^3(___) as the possible error. I'm having trouble filling in the blank. I'm not sure what dw, dx, dy, and dz would be. Any help?
     
    Last edited: Oct 24, 2004
  2. jcsd
  3. Oct 24, 2004 #2

    arildno

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    You truncate your numbers, right?
    So, the maximal error you make in the truncation of one number is 0.1.
    Agreed?
     
  4. Oct 24, 2004 #3
    the final correct answer is 25,000, which means the blank has to be .05. But I'm not sure where that number comes from. I also thought dw, dz, etc would be .1

    Edit: stupid mistake.. it would just be 50(.1)^3 to get the .05
     
  5. Oct 24, 2004 #4

    arildno

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    Looking closer, it says rounded. That means maximal error is 0.05 (do you see why?)
    Hence, you'd get:
    [tex]50^{3}(0.05+0.05+0.05+0.05)=25000*5*(0.2)=25000[/tex]
     
  6. Oct 24, 2004 #5
    could you give me a further explanation as to why the maximal error is .05?
     
  7. Oct 25, 2004 #6
    (50^3)(.05) + (50^3)(.05) + (50^3)(.05) + (50^3)(.05) = 25000

    The numbers are rounded....you cannot round to the first deciamal place and have the number and be off by more than .05

    Take 40.45 This would round to 40.5, with an error of .05
    Take 40.49 This would round to 40.5, with an error of .01
    Take 40.44 This would round to 40.4, with an error of .04

    Do you get it now?? You cannot round to the first decimal like that and be off by more than .05
     
  8. Oct 25, 2004 #7
    ah, I see thank you
     
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