mm-3759 === Subject: Re: Loan calculator equation, how do I go about it > Scenario: loan of 100,000 is made for 30 years at 5.75% interest per annum > and the repayments are 583.57 a month. > Can someone please tell me how to get that answer and how they got it.... You may get some help from the sci.math newsgroup's FAQ at http://www.cs.uwaterloo.ca/~alopez-o/math-faq/node76.html or from one of http://oakroadsystems.com/math/loan.htm Ken Pledger. === Subject: Re: equation On Tue, 21 Mar 2006 14:17:33 EST, mathmaniack (8^x-5^x)(7^x-2^x)(6^x-4^x)+(9^x-4^x)(8^x-3^x)(5^x-2^x)=105^x x=1 === Subject: Re: Cantor set and axiom of choice > in these matters, so could you please give me references > where these two things are proved rigorously? thank you Take your pick: 21,700 hits 494 hits 317 book pages Dave L. Renfro === Subject: Re: Cantor set and axiom of choice On Wed, 22 Mar 2006 14:20:09 EST, Modus.Ponens > Is it possible to construct the Cantor set C and to prove it > has the cardinality of |R without the axiom of choice? Yes. The classic middle-thirds Cantor set C is the set of all real numbers in [0, 1] that have ternary expansions that do not contain the digit 1. Clearly the identity map is an injection from C to [0, 1]. To go the other way, let x in [0, 1]. Then x has a unique non-terminating binary expansion. (E.g., 1/2 = 0.01111... .) In that expansion leave the 0's alone but replace each occurrence of the digit 1 with the digit 2, and interpret the result base 3. (So for 1/2 = 0.01111... we get 0.02222... = 1/3.) This map from [0, 1] is 1-1. The Schr.9ader-Bernstein Theorem now says that there's a bijection between [0, 1] and C, and the proof of the S-B Th'm. does not require the axiom of choice. [...] === Subject: Re: Cantor set and axiom of choice <10wfeyy4v8xas.xq643djm79i$.dlg@40tude.net Yes. The classic middle-thirds Cantor set C is > the set of all real numbers in [0, 1] that have > ternary expansions that do not contain the digit 1. You and Arturo Magidin no doubt know this, but for the benefit of others I thought I'd point out that there is a more elegant bisection argument that only makes use of the fact that the Cantor set is nonempty, closed, and dense-in-itself. In fact, this was the method Cantor originally used to prove that a nonempty perfect set has cardinality c, from which he was then able to conclude (via the Cantor-Bendixson theorem) that closed sets on the real number line are either countable or have cardinality c. Dave L. Renfro === Subject: Re: Cantor set and axiom of choice days. My association with the Department is that of an alumnus. >Is it possible to construct the Cantor set C and to prove it has the >cardinality of |R without the axiom of choice? I am not aware of anyone who constructs the Cantor set USING the Axiom of Choice, so the answer is yes. >If so , can anyone I would be far more interested in a reference that ->does<- use the Axiom of Choice. The usual construction is AC-free; the usual proof that it is uncountable is also AC-free (once you prove it consists of exactly those reals in [0,1] which have a ternary expansion that does not contain any 1's; or you can just ->define it<- that way. No AC involved). -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Cantor set and axiom of choice >>Is it possible to construct the Cantor set C and to prove it has the >>cardinality of |R without the axiom of choice? [...] > The usual construction is AC-free; the usual proof that it is > uncountable is also AC-free (once you prove it consists of exactly > those reals in [0,1] which have a ternary expansion that does not > contain any 1's; or you can just ->define it<- that way. No AC > involved). Note, though, that he was asking for more than uncountability: he wants a proof that |C| = |R|. For uncountability the usual diagonal argument does fine; for cardinality c you have to work just a little harder. Brian === Subject: Re: Cantor set and axiom of choice days. My association with the Department is that of an alumnus. >Is it possible to construct the Cantor set C and to prove it has the >cardinality of |R without the axiom of choice? >[...] >> The usual construction is AC-free; the usual proof that it is >> uncountable is also AC-free (once you prove it consists of exactly >> those reals in [0,1] which have a ternary expansion that does not >> contain any 1's; or you can just ->define it<- that way. No AC >> involved). >Note, though, that he was asking for more than uncountability: he >wants a proof that |C| = |R|. For uncountability the usual >diagonal argument does fine; for cardinality c you have to work >just a little harder. Okay: once you have that C consists of the reals in [0,1] whose ternary expansion does not contain any 1's, then there is an obvious injection from [0,1] to C by taking any number in [0,1], writing it in binary (omitting an expansion with a tail of 1's), and changing all the 1's to 2's. This is AC-free; the injection C-->[0,1] is of course AC-free, as is Cantor-Bernstein, which gives the equality. Little harder, for small values of little. (-: -- It's not denial. I'm just very selective about what I accept as reality. --- Calvin (Calvin and Hobbes) Arturo Magidin magidin@math.berkeley.edu === Subject: Re: Cantor set and axiom of choice >>Is it possible to construct the Cantor set C and to prove it has the >>cardinality of |R without the axiom of choice? >>[...] > The usual construction is AC-free; the usual proof that it is > uncountable is also AC-free (once you prove it consists of exactly > those reals in [0,1] which have a ternary expansion that does not > contain any 1's; or you can just ->define it<- that way. No AC > involved). >>Note, though, that he was asking for more than uncountability: he >>wants a proof that |C| = |R|. For uncountability the usual >>diagonal argument does fine; for cardinality c you have to work >>just a little harder. > Okay: once you have that C consists of the reals in [0,1] whose > ternary expansion does not contain any 1's, then there is an > obvious injection from [0,1] to C by taking any number in > [0,1], writing it in binary (omitting an expansion with a tail > of 1's), and changing all the 1's to 2's. This is AC-free; > the injection C-->[0,1] is of course AC-free, as is > Cantor-Bernstein, which gives the equality. Yep; exactly the same proof that I'd just given in my own response to the OP, except that I didn't leave out 1. ;-) (I chose the non-terminating expansions instead.) > Little harder, for small values of little. (-: Brian === Subject: Re: Limits at infinity >>LIM[x->oo](x^3-x^2-x+1)/(x^2+x+1) = >>= x^3(1 - 0 - 0 + 1)/(x^2(1 + 0 + 1))= >>= x >Where did the trailing +1 come from in your >numerator and denominator? >That certainly doesn't look right to me. Limit of f(x) = 1 is LIM[x->oo]1 = +1 === Subject: Re: Limits at infinity I guess I suck at trying to explain myself. What I meant to ask is how do we figure out how quotients of polynomials behave when going to infinity? In this particular case the degree of numenator was greater than that of denominator by 1. But what if its power was greater by two? How would it behave then when going to infinity? I just assumed it was important to know that but it most likely isn't. === Subject: Re: Limits at infinity > I guess I suck at trying to explain myself. > What I meant to ask is how do we figure out > how quotients of polynomials behave when > going to infinity? > In this particular case the degree of numenator was greater than that of denominator by 1. But what if its > power was greater by two? How would it behave then > when going to infinity? > I just assumed it was important to know that > but it most likely isn't. the highest power will dominate the expression just look at the highest powers on each side A_n*x^n A_(n-1)*X^(n-1) .... --------------------------------- B_m*x^m B_(m-1)*X^(m-1) .... tends towards A_n*x^n --------- B_m*x^m or (A_n/B_m) * x^(n-m) as X approaches infinity. -- Bye. Jasen === Subject: Re: Limits at infinity <3567359.1143067312792.JavaMail.jakarta@nitrogen.mathforum.org>, > I guess I suck at trying to explain myself. > What I meant to ask is how do we figure out > how quotients of polynomials behave when > going to infinity? > In this particular case the degree of numenator was greater than that of > denominator by 1. But what if its > power was greater by two? How would it behave then > when going to infinity? > I just assumed it was important to know that > but it most likely isn't. There are three cases: (1) degree of denominator > degree of numerator: limit zero (2) degree of denominator < degree of numerator: limit non-existent (3) degree of denominator = degree of numerator: limit is quotient of leading terms. With a bit of thought, it should be obvious why. === Subject: Re: Limits at infinity >I guess I suck at trying to explain myself. > What I meant to ask is how do we figure out > how quotients of polynomials behave when > going to infinity? Depends on which is dominent, the numerator or denominator. They both increase without bound, but one may increase quicker then the other. > In this particular case the degree of numenator was greater than that of > denominator by 1. Making it dominate the denominator, ie increase more quickly. > But what if its > power was greater by two? It increases even more quickly. > How would it behave then > when going to infinity? > I just assumed it was important to know that > but it most likely isn't. It *is* important to know, IMO. Taken separately, the numerator and denominator each approach infty as x approaches infty. In your case: lim x->oo x^3-x^2-x+1 = oo lim x->oo x^2+x+1 = oo I'll assume that's OK with you, but you're wondering how the limit of their *quotient* can be something other than infinite. Realize that as x grows larger and larger, x^3 will be a MUCH larger number than x^2. Try a few numerical examples. The lower powers and the constants are insignificant in this regard. What matters is the highest powers of x (generally called the degree of the polynomial.) Granted this is informal compared to other methods, but it is quite intuitive. When the numerator has higher degree than the denominator, it essentially reduces the problem to: lim x->oo x ...since x^3/x^2 reduces to x. Remember we are throwing out all lower powers of x and all constants. And of course that limit is infinite. IIRC you initially got as far as literally saying the limit was x. You just stopped prematurely, that's all. You needed to evaluate lim x->oo x, which is oo. Contrarily, when the denominator has higher degree the problem is essentially reduced to: lim x->oo 1/x ...same minimal ex. with degree 0 and degree 1 polynomials but this time the denominator has the higher degree. And that limit is 0. When they are the same degree, the problem reduces to: lim x->oo C/D *1 ...where C and D represent the leading coefficients. Remember x^n/x^n reduces to 1. This method is convenient for quickly evaluating limits of rational functions. It only takes a moment to look at: lim x->oo [20x^332 - 245x^21 + (369pi/4)] / [5x^332 + 35776x^331 - 6623549] ...and conclude the limit is 4. Can you see that? Both are of the same degree, 332, and C/D = 20/5 = 4. Neat. Alternatively, and *required* by some instructors (though I hope not yours!) you can algebraically manipulate the expression into a form where the limit can be determined. One general way of doing this is to divide both numerator and denominator by the highest power of x in the denominator. Try that way with a few examples readily available in your text under this section. Here's an ex: lim x->oo (2x-1)/(x+1) Divide thru by x (which is multiplying by 1/x) = lim x->oo (2x-1)(1/x) / (x+1)(1/x) = lim x->oo 2-(1/x) / 1+(1/x) = [lim x->oo 2 - lim x->oo 1/x] / [lim x->oo 1 + lim x->oo 1/x] ...applied properties of limits = (2 - 0) / (1 + 0) = 2/1 = 2 Now try that with the preceding limit :-) -- Darrell -- Darrell === Subject: Find zeros and maximum points-can't solve hello FIRST: -- How do I find range and zeros for function f(x) = sin[2x] - cos[2x] I'd say finding local maximum and minimum points on interval [0,Pi] and then comparing them would help me find the range of a function? I assume interval [0,Pi] is correct one since period is Pi and thus all other intervals have same max and min points? --How would I find zeros of a function? sin[2x] - cos[2x] = 0 sin[2x] = cos[2x] and then? Only thing I can think of is that since both cos and sin function have same value when angle is Pi/4 and since period is half the normal period ... functions would have same value at angle Pi/8 + k*2*Pi/2? Even if this is correct, it's done in a very brute force like way. Can it be done more elegantly? SECOND: -- Find zeros and local extremes for f(x) = x + 1 - 2*cos(x+Pi/3) And now what? thanx in advance === Subject: Re: Find zeros and maximum points-can't solve <13481211.1143074051341.JavaMail.jakarta@nitrogen.mathforum.org>, > hello > FIRST: > -- How do I find range and zeros for function > f(x) = sin[2x] - cos[2x] > I'd say finding local maximum and minimum > points on interval [0,Pi] and then comparing > them would help me find the range of a function? On any real interval of length at least Pi (one period) will work. > I assume interval [0,Pi] is correct one since > period is Pi and thus all other intervals have > same max and min points? > --How would I find zeros of a function? > sin[2x] - cos[2x] = 0 > sin[2x] = cos[2x] > and then? Only thing I can think of is that > since both cos and sin function have same > value when angle is Pi/4 and since period is > half the normal period ... functions would > have same value at angle Pi/8 + k*2*Pi/2? > Even if this is correct, it's done in a very brute > force like way. Can it be done more elegantly? On the interval 0 <= 2*x <= 2*pi there are two points at which sin(2*x) = cos(2*x), that at which sin(2*x) = cos(2*x) = sqrt(2)/2, and that at which sin(2*x) = cos(2*x) = -sqrt(2)/2. > SECOND: > -- Find zeros and local extremes for > f(x) = x + 1 - 2*cos(x+Pi/3) > And now what? For zeroes, solve f(x) = 0 For extrema start by solving f'(x) = 0? === Subject: Re: Find zeros and maximum points-can't solve On Wed, 22 Mar 2006 19:33:41 EST, newatthis >hello >FIRST: >-- How do I find range and zeros for function >f(x) = sin[2x] - cos[2x] >I'd say finding local maximum and minimum >points on interval [0,Pi] and then comparing >them would help me find the range of a function? >I assume interval [0,Pi] is correct one since >period is Pi and thus all other intervals have >same max and min points? >--How would I find zeros of a function? >sin[2x] - cos[2x] = 0 >sin[2x] = cos[2x] >and then? Only thing I can think of is that >since both cos and sin function have same >value when angle is Pi/4 and since period is >half the normal period ... functions would >have same value at angle Pi/8 + k*2*Pi/2? >Even if this is correct, it's done in a very brute >force like way. Can it be done more elegantly? Hint: Write it as: sqrt(2)* ( (1/sqrt(2))* sin(2x) - (1/sqrt(2))*cos(2x) ) and note that 1/sqrt(2) = sin(pi/4) = cos(pi/4). Then you might notice an addition formula you can use. --Lynn === Subject: Re: Find zeros and maximum points-can't solve THIS@asu.edu>: > On Wed, 22 Mar 2006 19:33:41 EST, newatthis >--How would I find zeros of a function? >sin[2x] - cos[2x] = 0 >sin[2x] = cos[2x] >and then? Divide both sides by cos(2x) -- legal since cos(2x) can't be zero in this equation -- and you should be able to solve the result quite easily. -- Stan Brown, Oak Road Systems, Tompkins County, New York, USA http://OakRoadSystems.com/ === Subject: Re: Mathematica, Maple, or Mathcad > I did a search on this and found > a similar question but it was from > 1997 so I thought I'd re-ask the > question and add some personal > details. > I am an undergrad math major also > studying physics for a minor. I > am looking to purchase and learn > Mathematica, Maple, or Mathcad. > So I am looking for personal experiences > and suggestions. Under the GNU license you can also attain the programs R and gnuplot. I know R isn't something a freshman should attempt firsthand, and most people would argue that it is mostly a statistical package, but I really like it for its intuitive structure and simplicity (plus it's free!). Gnuplot is a free package that helps you visualize any type of data (user defined or just plain old functions). Off topic: I ran into a built-in math visualization package on an old (and I mean old) Macintosh in a basement lab in my university just yesterday. It was simply beautiful. Why doesn't anybody put any effort into putting those kinds of programs on windoze? (has Apple gone that way too now?). My feeling: try all three (they all have trial versions, I'm sure). Go with your gut, although it would be wise to go with, as a previous poster mentioned, with the most common among your peers (the *ease* and M>