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Given that P=1.80 and Q=2.91, find ghe relative error in the computation of 1/P-1/Q

User VSe
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Answer:

±1.007%

Explanation:

You want the relative error in the computed value of 1/P -1/Q when P=1.80 and Q=2.91.

Maximum and minimum

Each of P and Q is specified to the hundredths place, so the errors in each of P and Q will be half of a hundredth, or 0.005. The maximum value of P or Q will be 0.005 more than the given value, and the minimum will be 0.005 less.

The maximum and minimum of the computation will be had when P and Q are at their opposite extremes: one is a maximum when the other is a minimum.

Max difference = 1/1.795 -1/2.915 ≈ 0.21404

Min difference = 1/1.805 -1/2.905 ≈ 0.20978

Relative error

The relative error in the computed value can be found a couple of ways. One is to compare the maximum and minimum computed values with the nominal computed value.

Here, we choose to compare half the difference in the computed values to the average of the computed values. That way, we can describe the error as symmetrical about that average;

((0.21404 -0.20978)/2) / ((0.21404 +0.20978)/2) = relative error

= (0.21404 -0.20978) / (0.21404 +0.20978) ≈ 0.01007

The relative error is about ±1.007%.

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Additional comment

Another way to compute the error is by looking at the derivatives. The relative error found that way is ...

RE = -∆P(Q/P(Q -P)) +∆Q(P/Q(Q -P))

where ∆P and ∆Q are ±0.005.

The maximum relative error found this way is about ±1.00686% compared to the ±1.00685% computed above.

If you compare the possible error to the nominal value, it is not symmetrical: -1.00532% to 1.00841%.

Looking at the above expression for RE, you see that the multiplier of the relative error in P is Q/(Q-P), while the multiplier of the relative error in Q is P/(Q-P). That is, errors in P have a much larger effect than errors in Q.

Given that P=1.80 and Q=2.91, find ghe relative error in the computation of 1/P-1/Q-example-1
User Dylan Jackson
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