Using Differentials to determine maximum possible error

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SUMMARY

The discussion focuses on estimating the maximum possible error in the product of four positive numbers, each rounded to the first decimal place, using differentials. The function defined is f(w,x,y,z) = wxyz, where w, x, y, z are all less than 30. The partial derivatives of the function are calculated as 1/2, reflecting the maximum rounding error of 0.5. The participants clarify the correct application of differentials and suggest using relative error for a more accurate assessment of the maximum possible error.

PREREQUISITES
  • Understanding of differential calculus
  • Familiarity with partial derivatives
  • Knowledge of error analysis in numerical computations
  • Basic concepts of rounding and its effects on calculations
NEXT STEPS
  • Study the application of differentials in multivariable functions
  • Learn about relative error and its significance in numerical analysis
  • Explore the concept of maximum error in rounding for various mathematical operations
  • Investigate the implications of rounding in real-world data processing scenarios
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Students in calculus, mathematicians, engineers, and anyone involved in numerical analysis or error estimation in computations.

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Homework Statement



Four positive numbers, each less than 30, 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.

So our function of four variables would be : f(w,x,y,z) = wxyz
Where w,x,y,z<30

Homework Equations



From Scratch

The Attempt at a Solution



So I have df = xyz\frac{\partial f}{\partial w} + wyz\frac{\partial f}{\partial x} + wxz\frac{\partial f}{\partial y} +wxy\frac{\partial f}{\partial z}

I know that

\frac{\partial f}{\partial w}, \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \&amp; \frac{\partial f}{\partial z} = \frac{1}{2}

since the max error when rounding a number is .5.

My question is when solving for the error, would I substitute w, x, y, \&amp; z with 30, or 29.9.

With 30 I get 303*(.2) = 5,400.
With 29.9 I get 29.93*(.2) =5,346.17 -> and I wouldn't know where to round off.

Or is my method wrong already?
 
Last edited:
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your differential isn't quite right
f=wxyz

then
df = xyz.dw + wyz.dx + wxz.dy + wxy.dz

a single partial derivative (excuse the notation) is
df/dw=xyz.1

and you know
dx,dy,dz,dw<=0.5
 
It might help to look at the "relative error"- that is, the error as a fraction or percentage of the actual value.
You have df= xyzdw+ xywdz+ xzwdy+ yzwdx. Dividing by the value, xyzw, gives
\frac{df}{xyzw}= \frac{dw}{w}+ \frac{dz}{z}+ \frac{dy}{y}+ \frac{dx}{x}

You know that x, y, z, and w are all less than 30 so f= xyzw< 810000. You also know that dcx, dy, dz, and dw less than 0.05.
 

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