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About Gauss

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  1. The order of the two hard disk drives on the one IDE ribbon cable does not matter, however one must be configured as master and the other must be configured as slave via jumpers (small connectors that fit over pairs of pins to program the drive through the hardware ) or by other means (cable select) If you have a master and slave on the primary IDE channel the first primary partition, on the master hard disk drive will be the "C:\" drive and the first primary partition on the slave hard disk drive will be the "D:" drive. This means that the master drive (on the primary channel IDE0) is the
  2. Yes, it's quite possible all you have to do is have one hard disk drive connected to the IDE0 slot and set it via the jumper to act as master. The other hard disk drive is connected to the IDE1 slot and set it via the jumper to act as master. You can have seperate independent multiple operating systems operating, one operating on the Primary master say Windows and the other say Linux on the Secondary master or vice versa. Then all you need is a boot loader program (Grub, Lilo, Boot Magic or System Commander come to mind) to launch the operating system you choose.
  3. When I read this a second time, when I made the first posting I thought he had a master slave, but he said Primary master and Secondary master which made me think along the lines of two independent drives. Because motherboards in all cases (generally) have two slots IDE0 and IDE1 for a total of four harddisk drives. We need more information to continue with this.
  4. If you use one IDE cable then the second hard disk drive has to be set to slave mode (using the jumpers). That is if you are running both harddisk drives on the one IDE cable. You also may use Fdisk in 'command mode' to ensure that the second drive is not set as an active drive.
  5. Nice equation, however there are 3 asymptotes, which yield 6 quadrants. Your equation adjusted and shown below would only fill the outer four quadrants. The inner quadrants, that is between x = 1 and x = 3 there is a couple of curves one convex and one concave. Do not asymptotically follow the asymptote lines. [math] f(x) = \left\{ \begin{gathered} \frac{1}{{e^{x^2 - 4x + 3} }} \hfill \\ \frac{{ - 1}}{{e^{x^2 - 4x + 3} }} + 2 \hfill \\ \end{gathered} \right. [/math] Similar sort of graphs are obtained when less than unity the outer quadrant graphs did not asymptotically fo
  6. The equation you are looking for is [math] f(x) = \left\{ \begin{gathered} \frac{1}{{x^2 - 4x + 3}} + 1 \hfill \\ \frac{{ - 1}}{{x^2 - 4x + 3}} + 1 \hfill \\ \end{gathered} \right. [/math] I hope this helps you
  7. You need to eliminate several possibilities as to why detection of the hard drives is inconsistant. Shut the computer down and open your computer casing and physically check to see if the 40gig drive has been set for the Primary IDE (master only) this is usually done by way of a jumper. You may need to get the details for this harddrive specification (from the net) to see if this is the case. Second check to see if your 10gig drive has been set for Secondary IDE (master only) this again is usually done by way of a jumper. You may need to get the details for this harddrive specification (
  8. The statement is not about a "finite decimal expansion". What is it that you do not understand about "for all of n in [math]\mathbb{N}[/math]. The set [math]\mathbb {N}[/math] has never been finite and will never be finite. Rationals are a subset of the reals. Therefore any number that is rational is also a real number. You may do that, but I will never deduce that. The proof states what it states. I agree with you unreservedly on this. It is very bad language to be used by any mathematician. I would also state that 0.999... expressed as such is also very bad l
  9. Discrete mathematics by its very nature, is simpler than the more traditional mathematics. The great breakthroughs in computer science has enhanced Discrete mathematics and given more impetus to this subject and has given us a better understanding of logic, number bases, algorithms graphs and Boolean algebra, predicate calculus and linear difference. These are just some of the subjects that were once thought of as esoteric subjects. Now, they are common place mathematical subjects that provide a fundamental understanding of our technological society and other mathematical subjects.
  10. Gauss


    The function f(x) = x is an identity function. When expressions have more than one operation, we follow rules for the order of operations. PEMDAS 1. Parentheses 2. Exponents 3. Multiplication and Division 4. Addition and Subtraction [math]f(x) = \sqrt{x^2}[/math] [math]f(x) = x^{\frac {2}{2}[/math] [math]f(x) = x[/math] Case when x = -3 [math]f(x) = \sqrt{-3^2}[/math] [math]f(x) = -3^{\frac {2}{2}[/math] [math]f(x) = -3[/math] Case when x = 3 [math]f(x) = \sqrt{3^2}[/math] [math]f(x) = 3^{\frac {2}{2}[/math] [math]f(x) = 3[/math] Another way Case
  11. Yes I agree with you that it is a "circular argument". Anybody can generate a polynomial function without knowing about Sin or the Taylor series (yes I know that everybody who gets to this stage should know about Sin). Therefore polynomial functions are continuous at x = c for each number c. So your statement "you are completely in the wrong" is wrong. The original problem also did not say that the Taylor series of trigonometric functions were not to be assumed to exist. Again a "circular argument".
  12. e(ho0n3 You state "This is a no no". I am sorry but you are wrong. I can and I did use the Taylor polynomial (series) with the explicit intention to show that sin(x) is continuous over its domain. Hence Sin(x) was shown to be continutous in Step 3 of the solution. If someone is just 'given' the Talyor polynomial then they can use basic limit theory for polynomials to show that Sin(x) is continuous without knowing about derivatives of functions which inturn automatically imply continuity. When you say "You destroy the purpose of this exercise by 'omitting' the proof". I totally d
  13. MandrakeRoot We can write Taylor polynominals: [math]P_n(x) = f^{0}(0) + f^{1}(0)x + \frac {f^{2}(0)}{2!}x^2 + . . . + \frac {f^{n}(0)}{2!}x^n[/math] in [math]\Sigma [/math] notation. If f is infinitely differentiable on an open interval I containing 0 then we have [math]P_n(x)= \sum\limits_{k = 0}^n {\frac{{f^{(k)}}}{{k!}}} x^k [/math] Hence we say that [math]f(x)= \sum\limits_{k = 0}^n {\frac{{f^{(k)}}}{{k!}}} x^k + R_{n+1}(x) [/math] and taking limits we say that f(x) can be expanded as a Taylor series in x and write [math]f(x)= \sum\limits_{k = 0}^{\infty}
  14. Gauss


    Reading your post I would assume that you mean the following First Part [math]f(x) = \sqrt {x^2}[/math] Hence [math]f(x) = x}[/math] Therefore x has the domain of [math](-\infty , \infty )[/math] Second Part The problem is to find the real zeros of a linear function f(x)=0 [math]f(x) = x^2 - 4[/math] [math](x -2)(x + 2) = 0[/math] [math]x = 2, x = -2[/math]
  15. Hanlin The other inverse fucntions are treated in a similar fashion. I hope this helps you.
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