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Round off error

f(x)=2x-sqrt(4x^2-1)

Under what circumstances could evaluating the function lead to round off error. How could you rewrite the formula to avoid this under those circumstances?

I have got no idea, Anyone would help? Thank you

Reply 1

Eau

what circumstances? when (4x^2 - 1) is not a square number?

Rewriting the formula will not change the function and so will not avoid the rounding off error.

Reply 2

Jonny W

If x is large then

2x . . . and . . . sqrt(4x^2 - 1)

are nearly equal. A computer is likely to round off each number to the same representation, and tell you that

2x - sqrt(4x^2 - 1)

is 0.

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You can help avoid the rounding error by computing the RHS of the identity

2x - sqrt(4x^2 - 1) = 1 / [2x + sqrt(4x^2 - 1)]

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This is the output of a C program I wrote to compute y = 2x - sqrt(4x^2 - 1) and z = 1 / [2x + sqrt(4x^2 - 1)] for five choices of x:

x = 1000, y = 0.0002500000155123416, z = 0.0002500000156250019
x = 1000000, y = 2.50060111284256e-07, z = 2.500000000000156e-07
x = 1000000000, y = 0, z = 2.5e-10
x = 1000000000000, y = 0, z = 2.5e-13
x = 1000000000000000, y = 0, z = 2.5e-16