Recent content by BvU

Yes, see #8: the hex representation of the smallest number > 2 is 40000001 I have muddied the waters by a practical interpretation -- makes for a good learning opportunity :smile: ##\ ##

The question is then: does the problem statement in #1 include this rounding (which is demonstrated in practice in the last block in #8, -- but may well be compiler-dependent)

So: no! Spring not stretched in the ##y## direction at ##t=0## doesn't mean never stretched. ##\dot y \ne 0## at t=0 means it does gets stretched in the ##y## direction. ##\ ##

In first instance this is math. You have two seond order differential equations, so four initial conditions. For example ##x, \dot x, y, \dot y## at ##t=0##. Or ##A, \phi_1, B, \phi_2## at ##t=0##. Setting leaves only one degree of freedom. And sure enough the "choice" ##x = A \cos(\omega _t),\...

I have shown that it is not the right answer (for my compiler). But probably the intended answer for a) is 2**(-23) Up to you :smile: !

So far I have done some trials starting with real*4 Y = 2.0 which is stored as 40000000 (in hexadecimal). Adding 2**(-23) (decimal 0.119209289551E-06) to this gets the sum stored as 40000000 again, so Y + 2**(-23) is NOT seen as greater than Y The minimum change required to see Y+...

Oops, messed up. Back later.

:welcome: The response of the system is simply a summation of the responses to the pulses from the past: ##V_0## at ##\tau_0##, so ##t-\tau_0## time ago, ##V_1## at ##\tau_1##, so ##t-\tau_1## time ago, etc. etc. Figure 2.7.3 is not very helpful in bringing this across. I expected a figure...

I thought. But it's more complicated. Trial and error, at least for me now. I figured the ##\varepsilon## for your part a) would be 2**(-22) but I find that anything > 2**(-23) already sets the last bit in that 40000000 with eps = 1.1920929E-07 ( 2**(-23), in hex: 34000000) I do Y = 2.0 + eps...

The key is in "real*4 2.0 is hexadecimal 40000000"

It means you have to reconsider your answer for a). real*4 2.0 is hexadecimal 40000000 2-23 = 1.1920929E-07 is hexadecimal 34000000 and 2+ 2-23 is hexadecimal 40000000 also. ##\mathtt {(2 − 2^{−23}) × 2^{127}≈ 3.4028235 × 10^{38}}## There are 23 bits for the fraction The first (implicit...

And please use a decimal point, not a decimal comma ... :rolleyes: What do you think of this: [edit] And I did ##\qquad##Y = 2 + ε ##\qquad##write (6,'(Z8.8)') Y-2.0 and it printed 00000000 ! ##\ ##

With a small sketch things become easier: In number 1 you have $$ F={G\,m_1m_2\over L^2} $$so in number 2 you want $$ {G\,m_1m_3\over x^2L^2}={G\,m_1m_2\over L^2}\quad\Rightarrow \quad m_2 = {m_3\over x^2} \quad \Rightarrow \quad m_1 = {m_3\over 2x^2} $$Then in number 3 you require $$...

Thanks, and I know that underneath a whole lot of newfangled stuff some sturdy fortran is actually doing the work. I also have done quite a lot of character stuff in fortran on VAX and Alpha. And their architecture handbooks (close enough to IEEE 754) are within reach right where I am sitting...

Great, thanks. Now: what do I have to do to resolve these LNK2019 errors ? I added /f77rtl in Project | Properties | Fortran as additional options - makes no difference Adding /f77rtl to the linker command line gives warning LNK4044: unrecognized option '/f77rtl'; ignored LINK...