Numerical Analysis Questions & Answers  
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.
09.17.17
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?
09.18.17
I'm not sure why you're getting negative bits.. You have more or less the right logic thus (although not perfectly). Keep working more on it.
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<b. Then the number of bits needed to store the exponent is 9. And maybe the number of bits needed to store the significand is zero. Zero is weird. In question it does not mention machine precision. Thus I think it can be zero. If I make mistake or miss something please forgive me. Please do not limit my message permissions.(I saw someone was limited in last year message.) Thank you very much.
09.19.17
I will answer your question if the numbers are properly typeset. Answer it again below and use to write the math and numbers properly.
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: $a<3*10^{-65},10^{32}<b$. Then the number of bits needed to store the exponent is 9. And maybe the number of bits needed to store the significand is zero. Zero is weird. In question it does not mention machine precision. Thus I think it can be zero. If I make mistake or miss something please forgive me. Please do not limit my message permissions.(I saw someone was limited in last year message.) Thank you very much.
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.
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