Question by Student 201627148 Professor,in denormal when e=0000 0000, 1.f change to 0.f. $$\#=(-1)^s*2^p*0.f$$ Here,if p=e-127, e=-127 not e=-126? Because e=0.But you wrote e=-126.
But $e=0$ indicates a special case, not an exponent. Thus, the minimum exponent $p_{\rm min}$ is not calculated using $e=0$ but using $e=1$.
 Question by Student 201427143 Professor, Sorry but I'm wondering about the assignment 5. The range upper limit is $10^{32}$, and the range lower limit is $3*10^{-65}$. So, I thought 2 equations. 1) $(-1)^s*2^{P_{max}}*1.f_{max}=10^{32}$ 2) $(-1)^s*2^{P_{min}}*1.f_{min}=3*10^{-65}$ Then, I got 2 size of bits 8 from eq1, and 9 from eq2. And I thought the minimum number of bits is 9. As at 5-(b) I put the 9 in equation $(-1)^s*2^{P_{min}}*0.f_{min}=3*10^{-65}$, I got the negative number of bits on significant. What is the problem with that?
 Question by Student 201529190 Dear Professor, in IEEE single precision format p=e-127. I know 127 come from $(2^8-2)/2$ (just like in the middle). In Assignment#1 Question#5, we are asked to minimize the number of the bits. When I caculate p=e-something, it is necessery for "something" need to be the number that in the middle? Just like 127 in single precision format? Or I can define this "something" whatever I want? (If it can change, maybe I can find a smaller number of the bits.) And I wonder if we need to consider the subnormal numbers in this question. In this question, if the "something" have to in the middle and we need not consider subnormal numbers. Then we just need to find a range (a,b), which: $a<3*10^{-65},10^{32} The number has to be constructed with the same rules as the float and double types shown in class. So yes, the exponents must be split equally between positive and negative values, and yes, the exponent must include the usual exceptions. Thus, denormal numbers will of course be present.  Question by Student 201427116 Professor, I have a question about Machine epsilon. We have used machine epsilon because of some possible errors on C language. In our class, for example, we calculated like this. $$g*g = (g\pm g*\epsilon_{mach})(g\pm g*\epsilon_{mach})$$ But in our textbook('Introduction to Numerical Methods', Jeffrey R.Chasnov) the author says like this: "A rough estimate would be$5(1+\epsilon_{mach})=5+5\epsilon_{mach}$, but this is not exact. The exact answer can be found by writing $$5 = 2^2(1+\frac{1}{4})$$ so that the next largest number is $$2^2(1+\frac{1}{4}+2^{-23}) = 5 + 2^{-21} = 5+4\epsilon_{mach}."$$ I can't understand why this differece is made between 5 and$2^2(1+\frac{1}{4}).$And also I'm wondering how can I express certain numbers to get rid of more errors as the author have done in our textbook.  09.24.17 Yes, the relative error can be less than$\epsilon_{\rm mach}$. In class,$\epsilon_{\rm mach}$was defined as the maximum relative error on a non-denormal float. For most non-denormal floats, the error will of course be less. 1 point bonus boost.  Previous 1 ... 6 , 7 , 8 , 9 Next • Make PDF $\pi\$