Gauss

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 one that gets booted first unless software intervens otherwise. As long as one drive is jumpered as master and the other as slave, any two IDE/ATA/ATAPI devices should work together. If you are using two drives on the same primary channel, it is important to ensure that they are jumpered correctly. Making both hard disk drives the master, or both the slave, will likely result in a very confused system.

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.

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.

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.

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 follow the asymptote lines but the inner quadrants did. However the most interesting value was when 0.999 was used [math] f(x) = \left\{ \begin{gathered} \frac{1}{{(0.999)^{x^2 - 4x + 3} }} \hfill \\ \frac{{ - 1}}{{(0.999)^{x^2 - 4x + 3} }} + 2 \hfill \\ \end{gathered} \right. [/math] We actually get the asymptotes y = 1, x = 1 and x = 3, the closer we get to Unity. Very interesting!

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

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 (from the net). This answer assumes that both harddisk drives are both connected to the motherboard seperately by way of seperate IDE cables. Boot up your computer and see if this fixes the first problem. If this does not solve the problem you need to shut down the computer and reboot, at the very beginning of the boot up sequence you need to enter the BIOS setup program (one method is to press Delete) when you see the computer display "Press DEL to enter setup" (or something similar) enter the BIOS setup program and go the section on 'boot up devices' this will have a list for floppys, cd, harddisks and other allowable boot devices. Check to see if the boot up sequence is listed correctly. This means that the computer must boot from the Primary master first (it must be listed as the first boot device if you want it to boot from it first). Then make sure that the Secondary master is listed as the second boot device. This should ensure that the computer boots from the Primary master first if this fails it will boot from Secondary master (assuming that it is allowed to boot) Exit out of the BIOS setup, ensuring you save the changes made when you exit. See how this goes. (It maybe more prudent to do the BIOS setup check first to see if this fixes the problem) I hope this helps in solving one of your problems

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 language (but thats my opinion). It is not rigorous enough, but everybody uses it that way, so by general agreement I also use it. Doing operations on this form leads to problems. Mathematics is only valid according to the definitions (and proofs) used and hence is only valid in that particular context in which it is used.

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.

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 when x = -3 [math]f(x) = \sqrt{-3^2}[/math] [math]f(x) = \sqrt{9}[/math] [math]f(x) = 3[/math] Case when x = 3 [math]f(x) = \sqrt{3^2}[/math] [math]f(x) = \sqrt{9}[/math] [math]f(x) = 3[/math]

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".

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 disagree. The problem was to show that inverse fucntions, especially the trigonometric ones are continous. This was done in the solution I provided, the question did not ask for proofs of basic limit theory. I agree when you say "You're doing it again", a proof would have been nice here, however I did say "I will keep it as brief as possible". However, I did leave some work for Hanlin to do. The defintion for Continuity at a Point states: Let [math]f[/math] be a function defined at least on an open interval [math](c - p, c + p) [/math] with p > 0. Then [math]f[/math] is continuous at c iff [math]\lim_{x \to \infty}f(x) = f© [/math] Therefore according to this defintion a fucntion is continuous at a point c iff: [math]f[/math]is defined at c. [math]\lim_{x \to c}f(x)[/math]exists. [math]\lim_{x \to c}f(x) = f© [/math] The defintion for Continuity on an interval states: Let (a, b) be an open interval. Then a function [math]f[/math]is continuous on (a, b) if it is continuous at each point c [math]\in[/math](a, b). This means that continuity at point can only be applied at that particular point. To prove that the function is continuous over the requried interval then you need to prove it by either step 3 or step 4 as was shown in the solution. Therefore your statement "but if the point is an arbitrary point in the domain of the function in question, then it follows that the function in question is continuous in its domain" is not correct.

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} {\frac{{f^{(k)}}}{{k!}}} x^k[/math] Therefore Taylor polynomials can be expressed as Taylor Series. You said Quote "But your step 3 involves changing two limits ! (that of the partial sums and that of x tending to c) which can be rather tricky...." Sorry no it does not. [math]\lim_{x \to \infty}P(x) =P©[/math] Is simply basic limit theory for polynomials.

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]

Hanlin The other inverse fucntions are treated in a similar fashion. I hope this helps you.

e(ho0n3 Your conditions only apply for continuity at a point.

Several steps are involved in this and I will keep it as brief as possible. Hanlin The 'Limit formula' you used is applied for only a point. Since continuity at a point c is defined in terms of a limit, there is an [math]\in ,\mbox{ }\delta [/math]version of the definition. A direct trnslation of [math]\lim_{x \to c} f(x) = f©[/math] One method of proving that arcsin(x) is continuous is as follows: Step 1 Let (a, b) be an open interval. Then a function [math]f[/math] is continuous on (a, b) if it is continuous at each point c [math]\in [/math] (a, b). Step 2 Sin(x) has a polynominal expansion derived by using a taylor series. [math]P(x) = Sin(x)[/math] [math]P(x) = x - \frac {x^3}{3!} + \frac {x^5}{5!} + ...[/math] Step 3 [math]\lim_{x \to c} P(x) = P©[/math] Proof omitted. In words, this result says that the limit of a polynomial P(x) as x approaches c not only exists, but is actaully the vlaue of P at c. Therefore polynomial functions are continuous at x = c for each number c. Step 4 [math]\lim_{x \to c} Sin(x) = Sin©[/math] Proof Take any number c. We can write [math]\lim_{x \to c} Sin(x) [/math] as [math]\lim_{h \to 0} Sin(c + h) [/math] [math]\lim_{h \to 0} Sin(c + h) = (Sin©) (\lim_{h \to 0}Cos(h)) + (Cos©)(\lim_{h \to 0}Sin(h))[/math] [math]\(Sin©)(1) + (Cos©)(0) = Sin© [/math] Step 5 Let f be a one to one function defined on an interval I. If [math]f[/math] is continuous, then its inverse [math]f^{-1}[/math] is also continuous. Proof omitted. [math]f(x) = Sin(x) [/math] The domain of [math]f[/math] is [math][-\pi/2, \pi/2][/math] [math]f'(x) = Cos(x) > 0[/math] for the domain of [math]f[/math]. Therefore [math]f(x) = Sin(x) [/math] has an inverse and therefore arcsin(x) is continous on the domain.

You are correct the symbol [math]\forall[/math] means "for all values of". The symbol [math] < [/math] means "less than" of course and adding a slant line to the less than symbol gives you the symbol [math]\leqslant[/math] which means "less than or equal to". The symbol [math]\geqslant[/math] means "greater than or equal to". I hope this helps you.

A friend comes along and gives you two numbers one is called [math]\alpha[/math] (alpha) and the other is called [math]\beta[/math] (beta). The friend also tells you that they (the numbers given to you) belong to a certain group of numbers called the real numbers and these numbers are represented by [math]\mathbb{R}[/math]. So you have these two numbers [math]\alpha[/math] and [math]\beta[/math]. Since you are a mathematical genius you want to arrange these two numbers in order. Then from the defintion of 'order of axioms' you can arrange these two numbers by looking at them and applying the following rules. a) [math] \alpha < \beta[/math] (This means the first number alpha is less than beta) b) [math] \alpha = \beta[/math](This means the first number alpha is equal to beta) c) [math] \alpha > \beta[/math](This means the first number alpha is greater than beta) Suppose your friend gave you these two numbers [math]\alpha = 3.78634 . . .[/math] and [math]\beta = 3.78629 . . .[/math]. Then from the rest of the lexicographical definition you can compare each single digit in the numbers [math]\alpha[/math] and [math]\beta[/math] and find out if one is less than, equal to or greater than the other. Of course you are very smart and you put the second number down first ([math]\beta[/math]) and then to the right and next to it you put down the first number ([math]\alpha[/math]). Because 3.78629 . . . is less than 3.78634 . . . I hope this helps you in understanding what is going on with all the mathematical terms.

From the ‘order of axioms’ in real number analysis, for every [math]\alpha , \beta \in \mathbb{R}[/math], exactly one and only one of the following holds. a) [math] \alpha < \beta[/math] b) [math] \alpha = \beta[/math] c) [math] \alpha > \beta[/math] Then an order for the real numbers can be laid down lexicographically. Let [math]\alpha \in \mathbb{R}[/math], and let [math]\alpha [/math] be expressed in the form: [math]\alpha = a_0.a_1a_2a_3 . . . a_k[/math] Let [math]\beta \in \mathbb{R}[/math], and let [math]\beta [/math] be expressed in the form: [math]\beta = b_0.b_1b_2b_3 . . . b_k[/math] When the first [math]a_k[/math] that differs from [math]b_k[/math] and [math]a_k[/math] < [math]b_k[/math] then [math]\alpha < \beta[/math], if [math]a_k[/math] = [math]b_k[/math] [math]\forall _k[/math] then [math]\alpha = \beta[/math] and if the first [math]a_k[/math] that differs from [math]b_k[/math] and [math]a_k[/math] > [math]b_k[/math] then [math]\alpha > \beta[/math]. Where [math]a_k[/math] are integers and [math]0 \leqslant a_k \leqslant 9[/math] and [math]b_k[/math] are integers and [math]0 \leqslant b_k \leqslant 9[/math] and [math]k = \mbox { }\{0, 1, 2, 3, . . .\mbox { }\}[/math] _____________________________________________________________ I would appreciate your feedback on improving (mathematically, more rigorous) the last four lines of the lexicographical definition.

Mandrake, how did you arrive at your solution?

For the case when [math] x \in \mathbb{C} [/math] Using two terms of the taylor series expansion we have [math] e^x - x = 0 [/math] [math] 1 + x + \frac {x^2}{2} - x = 0 [/math] [math] 1 + \frac {x^2}{2} = 0 [/math] [math]x_1 = (0 + i\sqrt {2})[/math] [math]x_2 = (0 - i\sqrt {2})[/math] Using three terms of the taylor series expansion we have [math] 1 + x + \frac {x^2}{2} + \frac {x^3}{6} - x = 0 [/math] [math]x_1 = (0.246017 + i1.28751)[/math] [math]x_2 = (0.246017 - i1.28751)[/math] [math]x_3 = -3.49203[/math] The number of solutions you require depends on how many terms you use in the taylor series expansion.

You are right Bloodhound I did not consider the case when [math]x \in \mathbb {C} [/math]

The problem is to find the real zeros of a nonlinear function f(x)=0. Let [math]y = e^x[/math] and [math]y = x[/math] Hence [math]e^x = x[/math] [math]e^x - x = 0[/math] Since the two function do not intersect there is no solution

Note the following pattern: [math] n = 0, \mbox { }then \mbox { } 1 - \frac {1} {10^0} = 0 [/math] [math] n = 1, \mbox { }then \mbox { } 1 - \frac {1} {10^1} = 0.9 [/math] [math] n = 2, \mbox { }then \mbox { } 1 - \frac {1} {10^2} = 0.99 [/math] [math] n = 3, \mbox { }then \mbox { } 1 - \frac {1} {10^3} = 0.999 [/math] [math] n = 4, \mbox { }then \mbox { } 1 - \frac {1} {10^4} = 0.9999 [/math] [math]\mbox{Then }1 - \frac {1} {10^n} = 0.9_19_29_39_4,...9_{n-1}9_n[/math] [math]Proof[/math] Let P(n) be the statement that [math] 1 - \frac {1} {10^n}[/math] has n 9's after the decimal place. [math]\forall_n \mbox { where } n \in \mathbb{N} [/math] [math] n = 0, P(0) = 1 - \frac {1} {10^0} = 0 \mbox {, thus P(0) is True.}[/math] [math] n = 1, P(1) = 1 - \frac {1} {10^1} = 0.9 \mbox {, thus P(1) is True.}[/math] [math] n = 2, P(2) = 1 - \frac {1} {10^2} = 0.99 \mbox {, thus P(2) is True.}[/math] [math] n = 3, P(3) = 1 - \frac {1} {10^3} = 0.999 \mbox {, thus P(3) is True.}[/math] [math] n = 4, P(4) = 1 - \frac {1} {10^4} = 0.9999 \mbox {, thus P(4) is True.}[/math] Assume that P(n) is true that [math] 1 - \frac {1} {10^n}[/math] (Basis of Induction) has n 9's after the decimal place. We shall show that P(n+1) is true. (Inductive Step) [math]P(n+1) = 1 - \frac {1} {10^{n+1}}[/math] [math]P(n+1) = P(n) + a_{n+1}[/math] [math]P(n+1) = 1 - \frac {1} {10^n} + \frac {10^1 - 1}{10^{n+1}}[/math] [math]P(n+1) = 1 - \frac {(10^n - 1)(10^{n+1})} {(10^n)(10^{n+1})} + \frac {(10^1 - 1)(10^n)}{(10^{n+1})(10^n)}[/math] [math]P(n+1) = 1 - \frac {(10^{2n+1} - 10^n)}{10^{2n+1}}[/math] [math]P(n+1) = 1 - \frac {1} {10^{n+1}}[/math] [math]\mbox{Therefore P(n+1) is True}[/math] [math]\mbox{Therefore P(n) is True}[/math] [math]\mbox{Therefore }1 - \frac {1} {10^n} = 0.9_19_29_39_4,...9_{n-1}9_n[/math] [math]\mbox{Therefore }1 = 0.9_19_29_39_4,...9_{n-1}9_n + \frac {1} {10^n}\mbox{ }\forall_n \mbox { where } n \in \mathbb{N} [/math]