I used y = y1 + y2 + y3 + .55 where I truncated at every step, and where the
intermediate variables are defined as y1 = x^3 - (x^3 % .001) = 2.571
y2 = -5 * (x^2 - (x^2 % .001)) = - 9.38 and y3 = 6 * x = 8.22
The sum was 1.9609999999999996. Truncating to 3 significant figures, 1.960.
The subtraction of the modulus is just a pythonista habit for truncation.
> Your answer, by the way, is correct, at 1.956853, but truncation at the end to 3
>significant figures would mean y = 1.956. Three figures of significance means 3
> after the decimal point. So it depends upon when you do the chopping.
>
> John (gnujohn) [Otherwise, a guy with an answer of 12.3 wouldn't have to work
> so hard, and his answer wouldn't mean as much, as the fellow who got .123 ]
I read "Use 3-digit arithmetic with chopping..." to mean that you chop on *every* operation. That would include each addition in the final sum, and not just one chop at the end of a sum of 3 or more numbers.
y=1.95685 (via plugging in the polynomial into wolfram alpha). You have already defined x=1.37, if you change the value of x then y will also have a different value.
((1.95685-1.95)/(1.95685))*100=0.35 so I get the same answer you do.
Evaluate the polynomial y=x^(3)-5x^(2)+6x+.55 at x=1.37.
Use 3-digit arithmetic with chopping (chopping means 'truncating' in this book). Evaluate the percent relative error.
My work:
y(1.37)=1.956853. This is the 'true val'
Chopping to 3 digits: 1.95=x
Percent relative error is ((true val-approxmation)/(true val))*100
==> Percent relative error is ((1.956853-1.95)/(1.956853))*100=.35 percent error
Is that correct?
Thanks so much! :D
