View Full Version : Differentials
circa415
Oct24-04, 01:15 PM
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?
arildno
Oct24-04, 01:41 PM
You truncate your numbers, right?
So, the maximal error you make in the truncation of one number is 0.1.
Agreed?
circa415
Oct24-04, 01:47 PM
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
arildno
Oct24-04, 02:09 PM
Looking closer, it says rounded. That means maximal error is 0.05 (do you see why?)
Hence, you'd get:
50^{3}(0.05+0.05+0.05+0.05)=25000*5*(0.2)=25000
circa415
Oct24-04, 10:22 PM
could you give me a further explanation as to why the maximal error is .05?
Spectre5
Oct25-04, 03:24 AM
(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
circa415
Oct25-04, 07:28 AM
ah, I see thank you
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