Question by Student 201327107 Professor, when you explained about float type machine precision you solved like this $$\epsilon_{machine}=\frac{(1*2^{-24})+1}{1}$$ But I don't know how does it derived. Could you explain how it derived
 09.20.16
No, this should read: $$\epsilon_{\rm mach}=\frac{1+2^{-24}-1}{1}$$ The first two terms on the numerator $1+2^{-24}$ correspond to the sum of the smallest number 1 and the largest possible round-off error $2^{-24}$. The last term on the numerator is the smallest number 1. I'll give you 0.5 point bonus boost.
 Question by Student 201527145 Professor, I have a question about an assignment #1_Question #2_(d). Does the smallest possible number mean the "positive" smallest possible number? Or should I consider the "negative" smallest possible number?
 09.21.16
Yes you are right: in Question A1Q2b and A1Q2d, we are seeking the smallest possible positive number. Thanks for pointing this out. I'll give you 2 points bonus boost.
 Question by Student 201327107 Professor, I don't understand about float type denormal number. I learned that denormal number condition is $$e=00000000$$then exponent should be $0-127$? But you explained exponent is $-126$. How comes?
The exponent of the denormal number is the same as the smallest exponent (-126) but the difference is with the significant which is 0.f instead of 1.f. Thus, the maximum denormal number is just below the minimum positive normal number. If the exponent would be -127, then there would be a large gap between the smallest positive normal number and the largest denormal number. Not a good thing! I liked your question, I'll give you 2 points bonus boost.
 Question by Student 201327102 Professor, I have a question. In double type, you taught us that $$p=e-g$$$$g=\frac{{e}_{max}-1}{2}$$ And you substitute $${2}^{11}-1(e={11111111111}_{base 2})$$ for $${e}_{max}$$ then in $${p}_{max}={e}_{max}-g$$ we should use different value in $${e}_{max}(for{11111111111}_{base 2}-1)$$ Although e=11111111111 is not possible(because it indicates infinite # or NaN), why we use e=11111111111 in former $${e}_{max}$$ 'after all'? What's the matter that we use e=11111111111-1 in former $${e}_{max}?$$
 09.22.16
I understand what is confusing you. When determining $g$, $e_\max$ refers to the maximum possible positive exponent. But in other cases it refers to the maximum possible positive exponent minus one (because the maximum positive exponent is reserved). They should have been written with 2 symbols in class to avoid confusion. In your notes, rewrite $e_\max$ to $e_\max^\prime$ when determining $p_\max$, with $e_\max^\prime=e_\max-1$. Good point, I'll give you 2 points bonus.
 Question by Student 201029134 professor, I think you should give some interval like a to b in The Question#1 on your assignment but you just give initial point $x_{0}$. and I think it is same in The Question#3 So, I wonder whether it is right or not.
 10.02.16
For A2Q3, there is no initial interval. For A2Q1, I have made a change to the question formulation. I'll give you 1.5 points bonus boost for pointing this out — I would have given more if you had not made spelling mistakes.
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