mm-2659 === Subject: Good (or not so good) online discrete math text I'm wondering if there is any good text online? Any suggestions please. === Subject: Question about the Master Thorem Consider the regularity condition a*f(n/b) .b2 c*f(n) for some constant c < 1, which is part of case satisfies all the conditions in case 3 of the master theorem except the regularity condition. === Subject: Re: Question about the Master Thorem Since you have chosen not to say what you mean by the master theorem, which is certainly not standard terminology, I doubt that many people will be able to answer this. === Subject: What is the algebraic manipulation required? A problem in Schaum's Differential and Integral Calculus transforms an equation as follows: |(4x^3 + 3x^2 - 24x + 22) - 5| = |4(x-1)^3 + 15x^2 - 36x + 21| = |(4(x-1)^3 +15(x-1)^2 - 6(x-1)| I can't figure out the rule/rules - the algebraic manipulations - that result in this transformation. Can anyone enlighten me? MMH === Subject: Re: What is the algebraic manipulation required? Well, it's easy enough to expand the last expression and see that it is equal to the first. So, maybe your question is: What would possess someone to do the algebra and how did they do it? It helps to know that the problem was to use the definition of limit to (4x^3 + 3x^2 - 24x + 22) - 5 vis a vis x - 1. Now, 1 is a root of (4x^3 + 3x^2 - 24x + 22) - 5, so x - 1 is a factor. Long division gives (4x^3 + 3x^2 - 24x + 22) - 5 = (x - 1)(4x^2 + 7x -17) = (x - 1)((4x^2 + 7x - 11) - 6) = (x - 1)((x - 1)(4x + 11) - 6) = (x - 1)((x -1)((4x - 4) + 15) - 6) = (x - 1)((x - 1)4(x - 1) + 15(x - 1)) - 6) = 4(x - 1)^3 + 15(x - 1)^2 - 6(x - 1). As a result, putting a bound on x - 1 puts a bound on (4x^3 + 3x^2 - 24x + 22) - 5 as desired. -- Paul Sperry Columbia, SC (USA) === Subject: Re: What is the algebraic manipulation required? |(4(x-1)^3 Parentheses are EVERYTHING in this exercise. :-) I should have spelled out the problem more completely and how insightful of you to know what I was talking about. MMH === Subject: T^2 + I = O Let V be an n-dimensional vector space over R (real field). Can we find a T in End(V) such that T^2 + I = O, where I is the identity transformation? Prove your idea! Thx. === Subject: Re: T^2 + I = O Let me rephase this. I have been given a homework problem and I am too lazy to try anything myself. Please give me the answer. I will give you a hint: there is a classical representation of the complex numbers as 2 by 2 matrices. === Subject: Re: Limit of a function-can't solve few problems Well who would have thought Dave, but boris actually doesn't post at that linky you gave. Nah, he posts here http://mathforum.org/kb/message.jspa?messageID=4149380&tstart=0 and here Rob Johnson's post is displayed as a single thread separated from one I started. === Subject: Re: Limit of a function-can't solve few problems Ah, you're welcome. In any case, I know that there is some way for you to signify the thread to which you are replying even using MathForum (since there is a References header to your last message). I read news using either Netscape's newsreader or Google and without a References header, they get confused as to where in the thread, and sometimes to which thread, a message belongs. These are just courtesies so that readers know to what your post refers. take out the trash before replying === Subject: 289^4 +578^4=4913^3 can anyone tell me how i figured it out? === Subject: Re: 289^4 +578^4=4913^3 ----------------- Zeez, if I understand correctly your question, you were asking how you figured out that or why: 289^4 +578^4 = 4913^3 Are you asking us to read your mind, or review in your mind what you did in figuring the above? Have you forgotten how you figured it? Tough question. I cannot read nor review minds, so let me just guess how you might have done it. Was it like this?: 289 is 17^2 = 17*17 289^4 is (289^3)*(289) 578^4 is (289*2)^4 = (289)^4 *(2^4) = (289^3)*(289)*(16) So, 289^4 +598^4 = (289^3)(289) +(289^3)(289)(16) = (289^3)[289 +289(16)] = (289^3)[(289)(1+16)] = (289^3)[(289)(17)] = (289^3)[(17^2)(17)] = (289^3)(17^3) = [(289)(17)]^3 = [4913]^3 = 4913^3 If it was not like that....sorry. === Subject: Re: 289^4 +578^4=4913^3 Or because it is in the form of a^4 + (2a)^4 we always get....the right side is x^3 where x = (17a^4)^(1/3) So if one has 1E6^4 + 2E6^4 we know it will be 2.57E8 (17 * (1E6)^4)^(1/3) And it is with some rounding error for a number that is not a perfect cube. === Subject: Re: x^2 + 2y^2 = w^2 This resembles an equation such as A^2+b^2+c^2 =d^2 where b and c are equal.If im reading you right ,lets reduce b and c to 1. If we assign a to be 2 the equation would be (2^2 -1-1)^2 + (2*1*2)^2 + (2*1*2)^2 = (2^2+1 +1)^2 If 3 is the value of a while b and c are both 1 one will get 7^2+6^2+6^2=11^2 When a is 4 , 14^2+8^2+8^2=18^2 . Does this help you? === Subject: Re: x^2 + 2y^2 = w^2 Because they both can't be even. === Subject: Re: x^2 + 2y^2 = w^2 They must be of the same sign: Suppose x is even, w odd, then x = 2k, w = 2m + 1 2|(4m^2 + 4m + 1 - 4k^2), which is impossible. Suppose w and x are even Then 2|(w^2 - x^2), so 2|2y^2, which is true. But y can't be even, for then (x,y,z) = 2, when it is 1. So y must be odd. However 2y^2 = 2 /= w^2 - x^2 = 0 (mod 4) So w and x must be odd. Suppose y is odd. Then 2y^2 = 2 /= 1 - 1 = 0 (mod 4) So, y must be even. Thus, w and x must be odd, y must be even. 2y^2 = w^2 - x^2 = (w + x)(w - x) (2) Case 2 | w + x. x,w odd; w - x even Since w - x is even, 2|(w - x) Case 2 | w - x. x,w odd; w + x even Since w + x is even, 2|(w + x) So in either case, 2 divides both factors, thus (w + x) (w - x) -------- and --------- 2 2 are integers. We next show that (w, x) = 1. We noted earlier that w and x must be odd, thus 2|x and 2|w. If some other number divides both w and x, it must divide y, but then (x,y,w) = 1, so this is not the case. We next show that (w + x) (w - x) ( -------- , --------- ) = 1 2 2 Let (w + x) (w - x) ( -------- , --------- ) = d 2 2 (w + x) (w - x) Then d|( -------- + --------- ), d|w 2 2 (w + x) (w - x) Also, d|( -------- - --------- ), d|x 2 2 But (w, x) = 1, so d = 1. Obviously, x^2 < x^2 + 2y^2 = w^2, so x < w, hence (w + x) (w - x) -------- and --------- 2 2 are positive. Now 2 2 2 (w + x) (w - x) w - x 2 y y 2 1 2 -------- X --------- = ------- = ----- = 2 ( --- ) = --- y 2 2 4 4 2 2 So What do I do with the 1/2 ? If I multiply it across, would, for example, (w - x) (w + x) and --------- 2 continue to be co-prime? -e === Subject: Re: x^2 + 2y^2 = w^2 Of course they have the same sign as you stated in (1.1) That is irrevalent to this part. Do you mean the same odd/even polarity? As (1) is homogenous, (x,y,w) = 1 iff any two of x,y or w are coprime. Not obvious, needs (1.1) Ok, perhaps useful point to make. Report it to HomeLand Security for being an evil thought. ;-) What's it? Why? One of them is already even, perhaps (w - x)/2. You have to be more picky than that. Perhaps even more crafty. Have you proved that theorem? Now since y is even, consider y^2 /2 = 2a^2 for some integer a and juggle it all around. === Subject: Re: x^2 + 2y^2 = w^2 Yes I meant that. Its in our book. 2 y 2 ---- = 2 a 2 2 2 y a = --- 4 2 y 2 2 2 2 2 (w + x) (w - x) w - x 2 y y 2 1 2 -------- X --------- = ------- = ----- = 2 ( --- ) = --- y 2 2 4 4 2 2 We need to find some way of showing y 2 2 2 ( --- ) = a , integer a 2 I'll have to think on this. -e === Subject: Re: x^2 + 2y^2 = w^2 We know that y is even, thus 2|y, so let y/2 = a, a^2 = y^2/4, 2a^2 = y^2/2, so: y 2 2 2 2 2 ( --- ) = 2 2 a = 4 a = (2a) 2 that? So, (w + x) (w - x) y 2 -------- X --------- = 2 ( --- ) = (2a)^2 2 2 2 So there are numbers u and v such that: (w + x) 2 (w - x) 2 -------- = u and --------- = v 2 2 So, (w + x) (w - x) 2 2 x = -------- - -------- = u - v 2 2 y = 2a (w + x) (w - x) 2 2 w = ------- + -------- = u + v 2 2 hmm, the back of the book says x = |d(u^2 - 2v^2)|, y = 2duv w = d(u^2 + 2v^2). -e === Subject: Re: x^2 + 2y^2 = w^2 Ok, close the book and prove it. No. (2a)^2 = y^2 Hm, it smells like dinner is almost ready. How similar to finding integer solutions for x^2 + y^2 = z^2. Have you done that? === Subject: Re: x^2 + 2y^2 = w^2 Suppose a and b are relatively prime positive integers and ab = c^n. We are to show that there are positive integers d and e such that a = d^n, b = e^n. Let the prime decomposititon of c be a1 a2 ak c = p p ... p 1 2 k na1 na2 nak Then c^n = p p ... p 1 2 k Since a and b are co-prime, it follows that their prime decomposition have none in common. na1 na2 nak ab = p p ... p 1 2 k naj So some p = a = d^n. Similarly for b. j Q.E.D. its in the book. -e === Subject: Re: x^2 + 2y^2 = w^2 === Subject: Re: x^2 + 2y^2 = w^2 coprime. Proof ok. Did you recall it or copy it from book? (w + x)/2 * (w - x)/2 = 1/2 * y^2 = 2a^2 (3.1) Hint. A factor on the left is even. ---- === Subject: Re: x^2 + 2y^2 = w^2 The d is from the fact that if: x^2 + y^2 = z^2, then it is true for all dx, dy, dz. So, after assuming (x,y,z) = 1, we may just tack on the d to each. How can we show this at my level. Lemma: if (x,y,z) = 1, x^2 + py^2 = z^2, some prime p, then any two of x, y, z are co-prime. Suppose (x,y,z) and x^2 + dy^2 = z^2. ??? -e === Subject: Re: x^2 + 2y^2 = w^2 remove each separably? Two cases depending upon which factor is even. Divide (3.1) by 2. Show a^2 = product two coprime factors. Can you take it from there? Have you forgotten the first equation? Elementary. For example If q prime and q | x,z, then q^2 | x^2, z^2, hence q^2 | py^2 and even if q = p, q | y I think for the lemma, all that's needed is for p to be square free. Suppose (x,y,z) what? === Subject: Dave Touretzky, the net-lifter net, informing about how the alleged free speech activist David Touretzky goes after publications that he doesn't like. But they were removed from the net. Why doesn't he say that the only free speech he stands for is just his own speech? BTW, his claim that I am obsessed with him is completely ridiculous. If I would have not been harassed with a porn letter about his sex toy purchases, which he ordered from his CMU office, I would probably never have researched him. All I want is that his wrongful activities are being corrected, and I gladly will forget that he exists. Barbara Schwarz -- http://www.thunderstar.net/~Schwarz/ (About Dave Touretzky) He also has bomb instructions on the net: http://www.religiousfreedomwatch.org/extremists/touretzky00.html How David Touretzky censored my website My name is Barbara Schwarz and for the past few years I have been attempting to expose racist statements made by a rat brain researcher at Carnegie Mellon University, David Touretzky. I have also been attempting to expose the fact this individual has purchased sexual implements using the university's phone number. Touretzky has systematically used his position as research scientist to intimidate every single ISP or web hosting to remove my website. Touretzky threatens with libel suits, but he never provided any evidence as to what the libelous statement was. Touretzky has never threatened to sue me for what I have made available on the web. Instead he goes behind my back and lies to ISPs and threatens them with a law suit. Why do you think he never sued me? Because he knows I have the documentation to prove what I write in my website. He knows that in discovery I will subpoena his telephone records, the places where he purchased his sex toys, using Carnegie Mellon facilities. Touretzky has consistently attacked my right to express my opinion. He has censored my website many times and is trying to silence me for telling the truth about him. Touretzky claims to be a free speech advocate, but not when someone is critical of him. For example, Touretzky is mirroring bomb information and promotes that website from his Carnegie Mellon University site. He uses a public chat room to make racist statements against African Americans, Hispanics, and other ethnic groups. Touretzky was confronted with his racist statements in a radio show and he did not deny them. Here are some of the racist statements David Touretzky made in a public chat forum: She [Diane Watson] is the former ambassador to Micronesia! and she's black. I should have known. What are all the really st00000pid congresswomen black? The black underclass is a very, very sick culture, with terrible internal problems such as high rates of black-on-black murder, teenage pregnancy, educational failure, drug abuse, and so on. AA programs can't help the black underclass, because those people are so screwed up they're virtually unemployable. ... ... they republished a long rant of mine about how women should have sole responsibility for conception ... a rant which was very well worded, and which I'm rather proud of now that they reminded me of it. There's also a long rant about the sickness of black underclass culture, which again, I completely stand behind. ... the Salvation Army ... buncha heads. Man, Hispanics are ed up, which is why they're still working class. Dips. the Muslims. .... check out the bloody Muslims .. jeez what loons. Hell, I'm for post-partum abortion! Retro-active abortion! Up until, say, the age of six months. Here is the site that Touretzky doesn't want you to see: PUBLIC SERVICE ANNOUNCEMENT To parents of students at Carnegie Mellon University If you are the parent of a student who attends Carnegie Mellon University, this PSA is to help you. David Touretzky is a research scientist at Carnegie Mellon University, and he teaches some of your children. David Touretzky owns a website that contains instructions on how to build bombs. One of these instructions encourages people to throw the bomb into police cars. Touretzky mirrored a site that was removed by the FBI after its original creator was arrested and prosecuted. David Touretzky claims he created the mirror as a matter of free speech. In my view this is not an innocent lack of judgment by David Touretzky for, as they say, where there is smoke there is fire, and Touretzky is involved in other activities. For example, during his work time he associates with individuals on the internet who harass religions and ethnic groups. Some of them posted threats against people they dislike. Touretzky has perversions that parents of CMU students should be informed about. He is a customer of sex shops and purchases sexual implements. The invoice posted on this website is the evidence of such activity, and it also shows that Touretzky used his University office phone number for the sex shop to contact him. You can compare the phone number listed in the invoice, 412 268-7561 (www.csd.cs.cmu.edu/people/faculty_s-z.html), with the phone number listed in the university directory which goes to David Touretzky's office. Touretzky has pornographic photographs on his CMU website, which he covered in part and removed some after I exposed him. There are some questions you should ask yourself when it comes to David Touretzky: Why would Touretzky provide instructions on how to build explosives? What if a student, like it happened in Columbine, downloads Touretzky's information, builds it and sets it off against people, who would be responsible? Why would the University allow Touretzky to use his office phone to order sex toys? Who is paying for Touretzky's activities when he does it from the University? Is it your tax dollar or tuition fees? There are deranged people and wannabe terrorists out there who want to know how to build explosive devices and use them against innocent victims. Even if Touretzky believes he has no responsibility for what he puts up on his website, he will eventually be accountable for the action that someone will undertake after having downloaded his bomb instructions and set it off. Let Carnegie Mellon University know about your views of David Touretzky. Barbara Schwarz === Subject: Re: Dave Touretzky, the net-lifter Could you explain what this has to do with mathematics? === Subject: Re: Dave Touretzky, the net-lifter Are you jealous he didn't use them on you? === Subject: Re: Dave Touretzky, the net-lifter Plonk. I was already sexually harassed by receiving the porn mail. You are primitive and perverted. Barbara Schwarz -- http://www.thunderstar.net/~Schwarz/ (About Dave Touretzky) Other interesting websites: http://www.religiousfreedomwatch.org/extremists/ http://www.alarmgermany.org/ http://bernie.cncfamily.com/sc/sitemap.htm http://www.cchr.org http://www.MindFreedom.ORG/ http://www.datafilter.com/mc http://www.freespeechstore.com http://www.amatterofjustice.org/amoj/00index.cfm === Subject: Re: Dave Touretzky, the net-lifter Barbara, where's the mathematics? Take your agenda elsewhere. We're not interested. Joe === Subject: equation help! could some one please help me rearrange for i? Will z=(1/(1+i)^n + r ((1- 1/(1+i)^n)/c+(1+i)^n/i)))/(1+i)^a/b)N === Subject: Re: equation help! Yes it is an interest rate problem. the equation is for the stettlement price onf a bond. I'm sorry that i could not write it correctly - please try this - ( ( 1 )) ( ( 1- ----- )) ( ( (1+i)^n )) ( 1 ( c + ---------)) ( -------- + r( i ))N z=( (1+i)^n ) (-----------------------------) ( (1+i)^a/b ) === Subject: Re: equation help! [[ It is always a good idea to include material from the post to which you are replying and also to reference the poster to whom you are replying. ]] (To which I responded - twice - pointing out unbalanced parentheses and then asking if this was an interest rate problem .To which Will responded:) I hope this is what you meant: Let I = 1/(1 + i); note I^n = 1/(1 + i)^n and if we can solve for I then we can solve for i. Solve for I: z/N = { I^n + r[ (1 - I^n) / (c + I^(n/i) ) ] } / I^(a/b) Providing I'm anywhere near right, I doubt there is a closed form solution and you will have to settle for a numerical approximation. Maple, Mathematica and, I suspect, Excel will do it to whatever degree of accuracy you need. There are participants in this newsgroup who are more proficient in the math of finance than I - maybe one of them can help. -- Paul Sperry Columbia, SC (USA) === Subject: Re: equation help! Is this an interest rate problem? If so, you might want to tell us what it is. -- Paul Sperry Columbia, SC (USA) === Subject: Re: equation help! Your parentheses don't balance - by my count 7 ( and 9 ). What do you mean by rearrange for i? Does (1 + i)^a/b mean (1 + i)^(a/b) or ((1 + i)^a)/b? Please put in some spaces: 1 + i not 1+i, etc.; especially ) ) ) not ))). It makes things easier to read. -- Paul Sperry Columbia, SC (USA) === Subject: Need advice for good book about combinatorics Any book you would recomend? Enumerative combinatorics is one book that I have heard good things about. === Subject: Re: Need advice for good book about combinatorics [1] Richard STANLEY , Enumerative Combinatorics, Woodsworth &Brooks/Cole Advanced Books&Software ,1986. [2] Ira M.GESSEL and Richard STANLEY , (Algebraic Enumeration,preprint ?}, try the homepages of the authors. COMBINATORICS: Topics,Techniques,Algorithms, Cambridge Univ.Press,1994 . Handbook of Combinatorics, vol. I-vol. II , Elsevier, Amsterdam,1995 . === Subject: (N+1)^N / N^N Last summer i had posted this statement.I still maintain that all values of N ,when not DECIMALIZED , are rational vulgar fractions. === Subject: Re: (N+1)^N / N^N It's not at all clear what you MEAN. You haven't said what N is- do you mean it to be a positive integer? If so, then clearly you don't mean all values of N ,when not DECIMALIZED , are rational vulgar fractions. since, obviously, N is an integer, NOT a vulgar fraction. I THINK you mean As long as N is a positive integer, then the (N+1)^N/(N^N) is a vulgar fraction. But that's clearly true. Certainly both (N+1)^N and (N^N) are integers and so (N+1)^N/(N^N) is a (vulgar) fraction, by definition. What is your point? === Subject: Re: (N+1)^N / N^N Show your work. === Subject: Re: (N+1)^N / N^N What statement does this statement refer to? Your thread title does not contain a statement. If this is important enough to you, you might want to spend more than 5 seconds with your post. For example, you didn't even put enough effort in your post to use correct period and comma spacing, which I find rather strange, because I think it's actually *harder* to mess up with these kinds of things -- the kinds things you would normally do without thinking. Did you actually try on purpose to not put a space after the period and after one of the commas? Dave L. Renfro (Trying to save you from William Elliot's wrath?) === Subject: Re: (N+1)^N / N^N Heh, *nobody* is safe from Williams wrath!! === Subject: Re: Derivatives-few questions thank you all for your help === Subject: Re: Derivatives-few questions When writting my last post I somehow forgot the difference between continuity and derivatives and a whole lot of confusion happened. Sorry for that Anyways,I get it in a way. But the whole point of derivatives is that with them we can estimate the slope of the graph at point in question,in other words where this point is going to go next.So why can't we with function f(x)=|x| approach zero from just the right side since point is obviously heading in that direction,and approaching zero from left just tells you the direction point or graph had prior to getting to x=0. It doesn't make a difference if we approach zero from left or right if line very close to point x=0 is linear, but since in this case it isn't linear it seems reasonable to assume graph has direction towards next point, which is point (x1,y1) where x1 is closest to x=0 and also x < x1 ? thank you for all your help === Subject: Re: Derivatives-few questions No. The derivative is the EXACT slope of a graph, not an estimate. You can say it! At the risk of beating a dead horse, let me again state what absolute value is: By definition: = -x if x < 0 What's not linear about that? Or alternatively, |x| = sqrt(x^2), but in this context, the first definition is the one you should focus on. This means, when considering the expression |x| in the definition of the derivative of f(x)=|x|, you have: f(x) = |x| For x any value less than zero, no matter how close to 0, we have |x|=-x by definition of absolute value. Furthermore, since h is as small a change in x as we desire, the expression |x+h| is also less than zero, thus |x+h|=-(x+h)=-x-h. For x any value greater than or equal to zero, |x| and |x+h| are x and x+h respectively. Evaluating f'(x) for both cases leads to two different values, -1 and 1. The limit therefore does not exist WHEN X=0, so f(x)=|x| is not differentiable at x=0. For x<0 the limit is -1. One way of stating the derivative of |x| is: d/dx[|x|] = |x|/x That's a reasonable statement on the condition you're talking about the derivative at any point OTHER THAN ZERO. At (0,0), which way should the direction of the graph be heading? Up, or down? (slope -1 or +1?) Wouldn't it seem reasonable to assume if you covered up the right side of the V the left side looks like a line (rather a ray) with slope -1? If you covered up the left part, would not the ride side have slope +1? I believe you are still confusing two different limits. There is no question that as x approaches 0 for |x| that the limit is 0. But it is NOT the case that the SLOPE of the curve approaches the same value from either side of 0. The curve _suddenly_ and very _impolitely_ (not well behaved) changes slope all the way from -1 to +1 immediately on either side of (0,0). It is not reasonable to say the SLOPE approaches the same value from either side of 0. That's why the derivative is undefined at x=0 for this function, despite the fact that the function itself IS defined at x=0. It's not only defined at x=0 it's continuous at x=0 (and everywhere else.) -- Darrell === Subject: Re: Derivatives-few questions You've lost me here, because |x| isn't linear in the linear algebra sense or in the elementary algebra sense of being an affine function. It is piecewise linear, however, but I'm not sure how this is relevant. What is relevant, and which you've pointed out, is that the two unilateral derivatives at x=0 differ. For the original poster, I'll mention that the question of whether a function is linear or not (or even piecewise linear or not) is not relevant because the function = -x^2 if x < 0 is differentiable at every point, including at x=0, and it's even easier to come up with similar examples that aren't differentiable at some point. (In fact, f'(x) is actually continuous, although the nice behavior ends about there, as the derivative of f'(x) doesn't exist at x=0.) Dave L. Renfro === Subject: Re: Derivatives-few questions I would think it is approximation since h is always different than zero. I look at graph as time...you can only travel forward.So if we approach zero from left it just shows the direction (slope) point had before x=0.But once x=0 it can't turn back to the left and since it can only move forwards it would seem reasonable to assume it has direction towards next point,and thus slope is calculated No! === Subject: Re: Derivatives-few questions h is nothing more than Delta x, a small change in x. This change in x approaches zero in the limit. It is this limit as h approaches zero that determines the derivative at zero. Which is -1. At x=0, that's right, since |x+h| and |x| are just x+h and x respectively once the absolute value bars are removed. But now, and this is the whole point, the value calculated is _different_ from the value calculated on the other side of zero, where |x+h|=-(x+h) and |x|=-x. When we're talking about a limit, they BOTH are considered. BTW it's not the fact that the graph can't turn back once x=0 is reached. Later you'll study parametric equations where graphs can indeed turn back. At this point you are just considering the graph as a whole in its xy Cartesian coordinates. It's because of the absolute value implications that -- Darrell === Subject: Re: Derivatives-few questions zero from left just tells you You could create two NEW definitions for the left-derivative and the right-derivative of a function at a given point. These would look only at approaching the point in question from the left or from the right, respectively. With these new definitions you could say f(x)=|x| has a left-derivative of -1 at x=0, and a right-derivative of +1. There's nothing wrong with this, and it may well be useful in certain situations... However, there are important results, like the Mean Value Theorem (MVT) which require the function to have a real (standard) derivative at every point in an interval for the result to hold. The MVT wouldn't work if we only require the function to have a left-derivative on the interval, or even if we required the function to have both a left and a right derivative on the interval! Your function f(x)=|x| would be a counter-example to such a modified MVT attempt. You could search on the web for Mean Value Theorem to get a better understanding of this if you're interested - the MVT doesn't use any complicated concepts that you've not met yet, so you should have no trouble understanding what it's saying, and seeing why it fails for f(x)=|x|, e.g. looking at the interval [-1, +1]. So whay you're suggesting isn't daft - it's just that for many (most?) applications, having only separate left- and right-derivatives is not a strong enough condition to be work with, so we (normally) concentrate on functions with full derivatives... Mike. === Subject: Simply confused hello There are functions that at particular points don't have a derivative. Examples of those function are :f(x)=|x| doesn't have derivative at point x=0 and f(x)=-|x-4|+5 doesn't have derivative at point x=4. I was wondering what are some tell-tale signs that particular function isn't differentiable at particular point? And I don't mean tell-tale signs when looking at a graph of a function, since from it you can quickly figure out whether function has such a point or not! About function f(x)=|x|.When trying to find derivative at point x=0 and approaching x=0 from the left I would assume the following would be the correct way to get derivative, but it is not! |x|=-x Instead,this is the correct way: But why is that? I understand if x was equal to -1 or some other negative number,then indeed |1|=-(-1).But since -x for x < 0 and so |0|=0. Therefor it really would make more correct way ! Confused to the max! thank you === Subject: Re: Simply confused I had another quick thought! My quick thought was that maybe you really were thinking of x being fixed as x=0 in the above formula (like you should), and you were just writing x instead of 0 with the assumption that x=0. (This is OK I guess). If x=0, then clearly x = -x = -(-x) etc. and there is no more to it than this! Both expressions are correct, and neither is more correct as they are the same. It's like arguing (pointlessly) whether 0 or -0 should be used. This distinction only makes sense if x != 0, and then it certainly is important which we use! === Subject: Re: Simply confused Yes,I was thinking of x being fixed as x=0. Due to my inability to express myself in coherent manner this thread has drifted a bit. I'd still need some tell-tale signs for when... === Subject: Re: Simply confused fixed as x=0 in the above formula (like you myself in coherent manner this thread Probably it would help if you think of this from the reverse angle first. There are loads and loads of functions you could easily come up with which are known to be differentiable at every point in their domain. You've learned that there's no problem f(x)=x^n, and there are several rules that allow you to combine two differentiable functions to get a new one. E.g. if f and g have a derivative, then their sum, product, difference and quotient (where it's defined) have a derivative. Also multiplying f by a constant will give another differentiable function, so straight away this means all polynomials have have derivatives everywhere. And quotients of two polynomials. And at some point you'll find that functions like sin(x), cos(x), e^x, log(x) have a derivative at each point in their domain. Finally the composition of two differentiable functions is also differentiable (recall the product rule). So in fact, you have to think a bit to come up with a regular sort of function which isn't differentiable, rather than the other way around! Using all the above rules for instance, you can easily see that the following bizarre looking functions are differentiable everywhere they're defined: f(x) = (x^3 + 17x + 44) / cos((sin(x^3))) f(x) = e^(tan(log(x)) etc. So what would trigger warning bells? If a function is discontinuous at a point, then it obviously can't have a derivative there, so this is a good starting point. Discontinous functions are often defined by splitting the definition up into different cases, so this is another alarm bell. Even when a function like f(x) = |x| (which is defined by two rules applying in different cases) is continuous, it may not have a derivative. Let's look at this more closely: |x| = -x (case 2, when x<0) This is like two separate functions joined together at x=0. E.g. if we consider only the domain x<0, then case 1 always applies, and in this domain f(x) is just a regular polynomial function. So you know f(x) has a will exist, so this only leaves the joining point at x=0 to worry about. Joining points you need to consider separately. I would think most functions you come across can be approached like this... Hope this helps, Mike. === Subject: Re: Simply confused You've just said you want the derivative at x=0? So you're trying to evaluate f'(0). Yes I suppose. It's true |0| = -0, but I can't help thinking you're already thinking something else. Regardless of whether this is right or wrong, you ought to be examining: Note x does not appear as you've already said x=0. (Why did you include x in your formula above?) OK, if we only consider h<0 , then |h| = -h, and the above limit would become = -1 But also we have = 1 Maybe, you actually want to evaluate the derivative of f at x, where we know x<0? This is different to what you set out to do, but is closer to what you evaluated above. So if you want to evaluate f'(x) where we know x<0, (e.g. maybe you want f'(-1) or f'(-1/1000))then we proceed as follows: = -1 [4] [1] is the definition of the derivative. [2] is applying the definition of f. Here you have to note that x<0, and also for small enough h also x+h<0. The limit is only dependent on this small h behaviour... [3] and [4] just simplify the value. So the above is showing that f'(x) exists for all x<0, and for these x, f'(x)=-1. But this doesn't mean f'(0)=-1 :) Aha! This is correct when evaluating the f' at a point x<0 as I did above. Now I'm confused about which point you're evaluating the derivative at. Clearly you're confused as well, so before going any further ask yourself what value of x am I trying to evaluate the derivative at?. Maybe you're confused about the variable that changes in the limit expression for the derivative? This variable is h, not x. So when we evaluate f'(x), we have and x is CONSTANT for the purposes of evaluating this limit! Once you've decided on your value of x, you will know how to evaluate f(x) and f(x+h) for small enough h, so it is routine to evaluate the limit... (just do the obvious substitutions) Mike. === Subject: Re: Simply confused Well || is a tell-tale sign. Watch your signs carefully; you stated yourself that |x| = -x for x < 0 ..and that... Well, f(x) was -x, so minus f(x) would be - (-x). Ohhhhh, I think I see what you're doing here. We're looking at what happens as h becomes infinitesimally close to 0. It's really important to know that h doesn't really hit 0; we are just making it as close as we please. Kyle === Subject: Re: Simply confused Well I think you've basically got the right idea. If you restrict x to values less than zero, the derivative -- the left-sided limit of the difference quotient -- is -1. And from the right, the limit is 1. Since the left and right limits don't agree, we say the limit doesn't exist. === Subject: Re: Simply confused I don't think he quite has the right idea. He knows what the answer is supposed to be, but thinks it would make more sense if it was something else. x]/h [f(x+h) - f(x)]/h. Now, if x < 0, then then f(x) = -x by definition of absolute value. So wouldn't this give a -(-x) term like you orginally I also want to comment that I don't think you're evaluating these left- and right-hand derivatives correctly. With the way you're doing it, you're actually doing two limits at once but kind of ignoring this fact. You're taking the limit as x approaches 0 from the left and taking the limit as h approaches 0. The better, and easier, way to do this is to plug in 0 for x, and let h approach 0 from the left and right sides With the way that you're doing it, it's not at all clear whether x + h is positive or negative, so I'm not sure that you're statement that f(x + h) = -(x + h) is accurate. With my way, you would get: h is approaching 0 from the negative side. For the right-hand derivative, you would get: Now this shows f(x) is not differentiable at 0, and I think it's a pretty clear and straightforward calculation. Mike === Subject: Re: Simply confused only if x=0 === Subject: Re: Simply confused Boris: |x| is a piecewise function.: =-x, for all x <0 This means, when evaluating the derivative in terms of x which is the usual way to get a derivative in terms of the VARIABLE x that can be used as a formula that spits out the slope of the curve at any x value where the slope is defined, you have to consider both cases. The definition of derivative is: For the second piece, we have |x|=-x, and also |x+h|=-(x+h) so this means: = -1 If x=0 as you suggest, then you use the other piece = 1 No offense, but you really need to understand these kind of things before going on to power rule proofs and what not. -- Darrell === Subject: Re: Gauss-Jordan Elimination Method Help! Subtracting R3 from R2, I end up with... [[1 -1 4 | 20] [0 2 4 | 20] [3 2 -1 | 3]] === Subject: Re: Gauss-Jordan Elimination Method Help! [[ It would be really nice to make posts self contained - it is unreasonable to expect responders to track down the post(s) to which you are referring. Also, it would be good to indicate to whom you are replying. ]] The problem was to reduce the matrix below to reduced row echelon form using only integer arithmetic. I use pseudo code ; R2 = R2 - R3 means the new Row 2 is the old Row 2 minus the old Row 3. Each step below refers to the result of the previous step. There are many ways to do this - almost all of them better than the one below. Starting with 2 5 -1 -3 1 -1 4 20 3 2 -1 3 (2) R2 = R2 - 2R1, R3 = R3 - 3R1; (3) R2 = R2 -R3; (4) R3 = R3 - 2R2; (5) R3 = R3 - 2R2; (6) R3 = (1/46)R3; (7) R1 = R1 - 4R3, R2 = R2 + 21R3; (8) R1 = R1 + R2. Gives 1 0 0 3 0 1 0 -1 0 0 1 4 Merry Christmas -- Paul Sperry Columbia, SC (USA) === Subject: Power rule proof-how did we get this? Hello two proofs I don't understand. 1-Proof that [(k*f)(x)]'=(k*f(x))'=k*f'(x): (k*f(x))'=k'* f(x)+k*f'(x)=k*f'(x) Huh?! How did we get k'*f(x) + k*f'(x)=k*f'(x)? 2-Power rule proof: if f(x)=x^n then f'(x)=n*x^(n-1) : I understand math induction proof so no need to explain it. In first step we prove that for n=1 f(x)=x^1, f'(x)=1*x^(1-1) = x^0 = 1 derive that P(n+1) is also true: f(x)=x^(n+1) f'(x)=( x^(n+1) )' = ( x * x^n )' = =1*x^n + x*n*x^(n-1) = x^n + n*x^n = =( n+1 )*x^n How did we get 1*x^n + x*n*x^(n-1) from ( x * x^n )' ? thank you === Subject: Re: Power rule proof-how did we get this? Here k is a _constant_. So what is k'? ************************ David C. Ullrich === Subject: orthonormal Jordan form (Reposted from wrong area) Does it suffice to norm the generalized eigenvectors as we assemble the Jordan basis, or would it be necessary to apply Gram-Shmidt after the fact? Does Gram-Shmidt preserve the Jordan matrix form? Alternatively (equivalently?), if we have a decomposition of a vector space V into dimV 1-dimensional invariant subspaces, can we pick vectors from each space such that the resulting basis of V is orthonormal? Any thoughts on this much appreciated! -Dylan Wright === Subject: Good tutor software What is some good software that will teach calc 1 and 2? I am looking for tutorial software and something I can plug in stuff to check work. Years ago I bought some stuff called mathmateca, are they even still around?