mm-446 Subject: Re: Looking for a good book on proofs...You might like:Proofs from the Book Trd EditionMartin Aigner, Gunter M. ZeiglerThe above book is full of satisfyingly short brilliant proofs that Erdos would consider to be in The Book.The Book in wch God maintains the perfect proofs for mathematicaltheorems, following the dictum of G. H. Hardy that there is no placefor ugly mathematics.Reading, Writing, ProvingA Closer Look at MathematicsUlrich Daepp, Pamela GorkinThe above book, wch is based on Polya's method of problem solving,aids students in their transition from calculus (or precalculus) togher-level mathematics.I would like to add:Inside CalculusGeorge R. ExnerSpringer Verlag 2000Though not strictly a profs book -- it presents the theoreticalpieces of introductory calculus, in a style suitable to accompanyalmost any first calculus text. === Subject: Re: Proving a(b - c) = ab - ac with the field axioms> If I have the field axioms:> .... > T3 b - a = b + (-a) That's really a definition (of the binary operation subtraction) rather than an axiom.> .... > And I'm asked to prove: a(b - c) = ab - ac> .... As an alternative to the suggestions by yourself and others, you could add ac to both sides. I tnk that method may be a bit shorter. . === Subject: Re: Proving a(b - c) = ab - ac with the field axioms Adjunct Assit Professor at the University of Montana.>If I have the field axioms:>(Ax1) x+y = y+x>(Ax2) x+(y+z) = (x+y)+z, x(yz) = (xy)z>(Ax3) x(y+z) = xy + xz>(Ax4) x + 0 = x, x * 1 = x>(Ax5) if x != 0 then there is a y such that xy = 1>And the following theorems at my disposal:>T1 if a+b = a+c, then b = c>T2 given a and b, there exists exactly one x such that a+x = b>T3 b - a = b + (-a)>T4 -(-a) = a>And I'm asked to prove: a(b - c) = ab - ac>How would I go about it?>So far i've done ts>a(b - c)>a(b + (-c)) (T3)>ab + a(-c) (Ax3)>I'm not sure how to prove a(-c) = -acYou want to prove that a(-c) is the unique element wch added to acis equal to 0.Lemma 1: 0a = 0 for all a. Proof. 0 + 0a = 0a = (0+0)a = 0a + 0a;Lemma 2: If x+y = 0, then y=-x. Proof: x+y = 0 = x + (-x); from T1, y = -x. QEDProp. 1: a(-c) = (-a)c = -(ac). Proof: ac + a(-c) = a(c+(-c)) = a(0) = 0. By Lemma 2, a(-c) = -(ac). ac + (-a)c = (a+(-a))c = 0c = 0. By Lemma 2, (-a)c = -(ac) = a(-c). QEDNow, for practice, prove that (-a)(-c) = ac.-- === Subject: I got the answer myslef! by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2F3I0H20089;I got the answer myslef!f(x)=g(x) : 2p(1-p)f(x!=g(x) : p*p+(1-p)*(1-p).P(f(x)=g(x))=P(f=1,g=1)+P(f=0,g=0)=P(f=1|g=1)P (g=1)+P(f=0|g=0)P(g=0)f and g are independent，thereforp(f=1|g=1)=P(f=1)，P(f=0|g=0) =P(f=0)so the answer is: p(f=g)=2p(1-p).am I right? === Subject: Radical division by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2F3I1O20145; guys, I'm working through my Calculus book and I came upon aexample to find the limit of a function by using the limit laws. Thelaws aren't a problem but The example involves a rational functionthat has a redical in the numerator:(2x^2 + 1)^1/2 / 3x - 5The example goes on to divide both the numerator and denominator by x,but ts is where I can't continue working with the example... becauseI can't figure out how (2x^2 + 1)^1/2 = (2 + 1/x^2)^1/2Can anyone give me nt or briefly explain how the numbers move toyield that result?Best regards,Ivan === Subject: Re: Radical division>The example goes on to divide both the numerator and denominator by x,>but ts is where I can't continue working with the example... because>I can't figure out how (2x^2 + 1)^1/2 = (2 + 1/x^2)^1/2Not quite. Rather, [ (2x^2 + 1)^1/2 ] / x = (2 + 1/x^2)^1/2Here's how: [ (2x^2 + 1)^(1/2) ] / x [ (2x^2 + 1)^(1/2) ] / [ (x^2)^(1/2) ] [ (2x^2 + 1) / x^2 ]^(1/2) [ 2 + 1/x^2 ]^(1/2) === Subject: Re: Radical division by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2FBLqq32507;> guys, I'm working through my Calculus book and I came upon a>example to find the limit of a function by using the limit laws. The>laws aren't a problem but The example involves a rational function>that has a redical in the numerator:>(2x^2 + 1)^1/2 / 3x - 5>The example goes on to divide both the numerator and denominator byx,>but ts is where I can't continue working with the example...because>I can't figure out how (2x^2 + 1)^1/2 = (2 + 1/x^2)^1/2>Can anyone give me nt or briefly explain how the numbers move to>yield that result?>Best regards,>Ivan Ivan :Assuming that you want the limit as x -> + oo write(2x^2+1)^(1/2) = x*(2+1/x^2) , 3x-5 = x(3-5/x) =>(2x^2+1)^(1/2)/(3x-x) = (2+1/x^2)^(1/2)/(3-5/x).Hence the limit is sqrt(2)/3. === Subject: Re: Radical division> The example involves a rational function> that has a redical in the numerator:> (2x^2 + 1)^1/2 / 3x - 5 [ (2x^2 + 1)^1/2 / 3x ] - 5Is that what you mean?> The example goes on to divide both the numerator and denominator by x,> but ts is where I can't continue working with the example... because> I can't figure out how (2x^2 + 1)^1/2 = (2 + 1/x^2)^1/2You write bogus. That's not an identity. Do you mean (2x^2 + 1)^1/2 / x = (2 + 1/x^2)^(1/2) ?Then you can use a sqr b = (sqr a^2) sqr b = sqr (a^2 b)> Can anyone give me nt or briefly explain how the numbers move to> yield that result?Take the necessary time and care to accurately present formulas. === Subject: Re: Radical division> (2x^2 + 1)^1/2 / 3x - 5The example goes on to divide both the numerator and denominator by x,> but ts is where I can't continue working with the example... because> I can't figure out how (2x^2 + 1)^1/2 = (2 + 1/x^2)^1/2You mean, you can't figure out why(2x^2 + 1)^1/2 / x = (2 + 1/x^2)^1/2and it's because(2x^2 + 1)^1/2 / x =(2x^2 + 1)^1/2 / (x^2)^1/2 =((2x^2 + 1) / (x^2))^1/2 =(2 + 1/x^2)^1/2-- If ts message helped you, consider buying an itemfrom my wish list: === Subject: help by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2F3I4q20239;Help!! can someone help me with the name of or how to find the puzzledrawings. I don't know how to explain them other than the mexicanriding a bike, all you can see is a few circles and lines. Another isa bear climbing a tree, and an all wte page is a Polar Bear in asnow storm. === Subject: Re: Intergrals by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2FBLqh32476;> yogi..>my lecture notes didnt teach us wat trig substitutions to sub.. so im>quite blur but using trig to sub.. hee hee.. so im learning all i can>in ts forum.>i've another 2 qns here.. (probably involving trig sub as well)>Int x/(1+9x^4) dxSubstitute y = x^2.You'll arrive at 1/2* Int dy/(1+9y^2) wch is easy to integrate if you know that Int dy/(1+y^2) = arctan(y) + c.>Int (2x^2+11)/(x^2+4) dxPolynomial division gives (2x^2+11)/(x^2+4) = 2 + 3/(x^2+4).Now integrate both terms separately and use again that Int dy/(1+y^2) = arctan(y) + c.Best wishesTorsten. === Subject: Re: Intergrals> my lecture notes didnt teach us wat trig substitutions to sub.. so im> quite blur but using trig to sub.. hee hee.. so im learning all i can> in ts forum.I don't know if you've learned partial fractions yet, but chances are that you will soon or already have. If you haven't, you may find useful and interesting. Whether you know partial fractions or not,though, the important point about partial fractions is ts:Using partial fractions, any integral of the formInt P(x)/Q(x) dx, where P and Q are polynomials in x can be reduced to an integral of a sum of the following:1. polynomials in x2. A/(Bx + C)^D3. (Ax + B)/(Cx^2 + Dx + E)^Fwhere A, B, C, D, E, and F are conts.Therefore, you can integrate any ratio of two polynomials if you can integrate the above three kinds of expressions. You know how to integrate polynomials, you know how to integrate A/(BX + C)^D, and you only ever need two kinds of trig substitution to integrate (Ax + B)/(Cx^2 + Dx + E)^F.So, what you need to remember is that whenever you see sometng of the form (Ax + B)/(Cx^2 + Dx + E)^F, you will always use the same trig substitution.Here's what you doFirst, you apply partial fractions. After you are done, if you have any terms of the form (Ax + B)/(Cx^2 + Dx + E)^F:1. Eliminate the Dx term with a substitution y = sqrt(C) x + D/(2 sqrt(C))Now you have sometng of the form (Gy + H)/(Iy^2 + J)^F2. Divide numerator and denominator by J^KNow you have sometng of the form (Ky + L)/(My^2 + 1)^F3. Apply the trig substitution to eliminate the denominator: y = tan(z) / sqrt(M)The reason that ts trig substitution is useful is that tan^2(z) + 1 = 1 / cos^2(z) -- i.e., it transforms the denominator into a part of the numerator, as you will see when you solve a few problems ts wayNow you have sometng of the formN tan(z) cos^Q(z) + P cos^R(z)wch can be integrated easily (left half using the substitution u = cos z, right half by parts).> i've another 2 qns here.. (probably involving trig sub as well)> Int x/(1+9x^4) dxYou want the integral to be a linear term divided by a quadratic term. However, you have a 4th power here, so first you substitute y = x^2, and you get 1/2 1/(1 + 9y^2) dyNow You have sometng in the desired form (linear divided by a quadratic), and the quadratic denominator doesn't have a term in y, and the cont in the numerator is already 1, so you can move directly to the trig substitution.You want the denominator to become tan^2(z) + 1. Therefore, you want 9y^2 to be tan^2(z), i.e. 3y = tan(z), i.e.:y = tan(z)/3dy = 1/3 1/cos^2(z) dz (by quotient rule)Int 1/2 1/(1/cos^2(z)) * 1/3 1/cos^2(z) dz =Int 1/6 dz =1/6 z + C =1/6 arc tan (3y) + C = 1/6 arc tan (3x^2) + C> Int (2x^2+11)/(x^2+4) dxHere your denominator is in a useful form (quadratic), but the numerator isn't. You can apply partial fractions or simply see that 2x^2 + 11 = 2 (x^2 + 4) + 3, so that you actaully have:Int (2 + 3/(x^2 + 4)) dxat wch point you can proceed by dividing the numerator and the denominator by 4 and applying a trig substitution. -- === Subject: Greatest integer functionI would like to show f(x)=[x] is continuous on every noninteger x in R, where [x] denote the greatest integer less than or equal to x.By definiton of continuoity, let c in R-Z,For any e>0, If there exist d>0 such that |x-c| |[x] - [c]| I would like to show f(x)=[x] is continuous on every noninteger x in R> , where [x] denote the greatest integer less than or equal to x.> By definiton of continuoity,> let c in R-Z,> For any e>0,> If there exist d>0 such that |x-c| |[x] - [c]| Then ts imply f(x)=[x] is continuous on every noninteger x in R> but,> I don't know how do I put above d=?Huh? Don't underd. You want value for d? d = min(c - [c], [c]+1 - c) === Subject: Can anyone slove ts?I have an equation for Vout given AH (absolute humidity in g/m3) and T(temperature in Deg. C.) as followsVout = (a * AH * AH + b * AH) * (c * T * T + d * T + e) * fConts: a = -0.00067742 b = 0.17704445 c = -0.000017156 d = -0.00088115 e= 1.11463 f = 1.062806I need it worked to give an equation for AH given Vout and T. === Subject: Re: Can anyone slove ts?>I have an equation for Vout given AH (absolute humidity in g/m3) and T>(temperature in Deg. C.) as follows>Vout = (a * AH * AH + b * AH) * (c * T * T + d * T + e) * f>Conts: a = -0.00067742 b = 0.17704445 c = -0.000017156 d = -0.00088115 e>= 1.11463 f = 1.062806File these away for a little wle.>I need it worked to give an equation for AH given Vout and T.The technique is the same as always: you need to get AH on one side by itself, with no AH on the other side. I'll begin by formatting your equation in a more dard way: Vout = (a*AH^2 + b*AH) * (c*T^2 + d*T + e) * fNow, what is keeping AH from being alone on one side? It's got some multiplications (by a and b) and additions going on, and then the result is multiplied by (c*T^2 + d*T + e)*f. In solving an equation you work from the outside in; you reverse the order of operations. To undo a multiplication you need a division, so divide both sides by (c*T^2 + d*T + e)*f: Vout / [(c*T^2 + d*T + e)*f] = a*AH^2 + b*AHNow ts is revealed as a garden-variety quadratic. Subtract the LHS from both: 0 = a*AH^2 + b*AH - Vout / [(c*T^2 + d*T + e)*f]and if you like you can reverse the LHS and RHS to make ts equation look more like dard form: a*AH^2 + b*AH - Vout / [(c*T^2 + d*T + e)*f] = 0Now solve for AH: AH = [ -b +- sqrt(b^2-4a*Vout/[(c*T^2+d*T+e)*f]) ] / (2*a)Now plug in the specific numbers you gave earlier, and do any possible simplifying.-- === Subject: Re: Can anyone slove ts?> I have an equation for Vout given AH (absolute humidity in g/m3) and T> (temperature in Deg. C.) as follows> Vout = (a * AH * AH + b * AH) * (c * T * T + d * T + e) * f> Conts: a = -0.00067742 b = 0.17704445 c = -0.000017156 d = -0.00088115e> = 1.11463 f = 1.062806> I need it worked to give an equation for AH given Vout and T.> nt: You will eventually have to solve a quadratic in AH. You needto rearrange the equation and then decide wch branch to keep.-- === Subject: Re: Can anyone slove ts?> I have an equation for Vout given AH (absolute humidity in g/m3) and T> (temperature in Deg. C.) as follows> Vout = (a * AH * AH + b * AH) * (c * T * T + d * T + e) * f> Conts: a = -0.00067742 b = 0.17704445 c = -0.000017156 d = -0.00088115 e> = 1.11463 f = 1.062806> I need it worked to give an equation for AH given Vout and T.> Is'nt ts just a quadratic in AH with somewhat messy terms? === Subject: Math, Avoiding burnout, study tips recommended:Second, I'm looking for advice or how some people study to avoidburnout. I tnk Math is one of the most demanding disciplines in termsof the amount of study/practice time required to get a good grade in thecourse. Whereas one can get good grades in other types of classes bynaturally/quickly underding the concepts and not spending hoursweekly on studying, I don't feel the same applies to math (at least the200-level courses (calculus, linear algebra, etc..)... I don't feelthere's any shortcuts to getting good grades in math classes. The onlytng one can do is to spend the hours weekly practicing, practicing andpracticing even more.However, I wonder if I'm overstudying? For 2 months, I spend about2-3 hours per day. Sometimes 4 hours. I have not taken one day off yetof study. I'm starting to feel a little burnout since I have 2 monthsmore to go until the finals. I tnk overstudying can be detrimentalas well. What I tend to do is to study ahead on material that theteacher has not covered yet. I try to underd that as much as I canso that when the teacher goes over it, I'm hoping they'll fill in thegaps (wch are the concepts I did not underd wle studying ahead).Just wondering how some folks approach their time-management/study fortheir math classes? === Subject: Re: Math, Avoiding burnout, study tips recommendedI forgot to mention:I'm a firm believer in studying a little bit everyday (even if it's just1-2 hours) rather than taking 2-3 days off at a time and cramming a6-hour study session. The approach of spending 1-3 hours daily and notdoing *any* overnighter study cram-sessions has gotten me good gradesover the years. The only question there is if it's beneficial to take at least 1 day offper week ? Right now, that's sometng I have not done. === Subject: Re: Math, Avoiding burnout, study tips recommended>I forgot to mention:>I'm a firm believer in studying a little bit everyday (even if it's just>1-2 hours) rather than taking 2-3 days off at a time and cramming a>6-hour study session. The approach of spending 1-3 hours daily and not>doing *any* overnighter study cram-sessions has gotten me good grades>over the years. And I tnk it should work for most students.I truly believe that for most math courses up through the undergraduate level math ability is relatively unimportant for success. What matters much more is good work habits, especially time management.Nobody would try to prepare for a big game by sitting around eating potato cps for six days and then practicing several hours the night before, yet many students approach math that way. Talk about shooting oneself in the foot!>The only question there is if it's beneficial to take at least 1 day off>per week ? Right now, that's sometng I have not done.Everybody is different. I would say if you have a regime that works for you, you should not be too quick to change it.-- === Subject: Beware undergrads, copiers of your postsJust so you know, there are people who might copy your posts and putthem up on a webpage!Considerhttp://homepages.cwi.nl/~dik/english/ mathematics/jsh.htmland the post he copies is YEARS old, so don't tnk that posts youmake here might not come back later, on a webpage, used by people whodon't like you.Sure, posts are arcved anyway, but people have to dig for those.Here if you notice by doing a Google search on my name and math thatDik Winter's webpage comes up prominently.He's stolen my writings and used them to promote mself, banking onthe attention that my work garners, and when I requested that heremove them, he refused.He lives in the Netherlands. I guess they don't care about copyrightsthere, wch surprises me since it *is* a European nation, but thenEurope isn't what it used to be.Then again, maybe he isn't so safe in the Netherlands, but he sure isacting like he's safe.In any event, you're fair warned. You can look here to see what couldhappen to you. === Subject: Re: Math Help! *Testing* by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2FG2uD04347;>What is 2+2?>Call 4 Hobbit * Testing 1-2-3 *The answer is 3. === Subject: Re: matrices of revolution>i was trying to figure the dard matrix for rotating a vector>around another vector some number of radians in 3-D and i cant find>any where in my linear algebra a way to do it. ive found a few>different little 'nts' on how to do it i tnk. like if u find e1 e2>and e3 (basically i j and k) and find what they become after the>rotation then the dard matrix will be [a1 a2 a3] where a1 a2 and>a3 are the new transformed vectors. but i have no idea how to find a1>a2 and a3. also, ive found how to do it around an axis. but im still>very confused on how to do it around any random vector.>any help would be niceLook at the picture here:http://mathworld.wolfram.com/RotationFormula.htmlThe vector 'ON' is the axis, OP is the unrotated vector, and OQ is thevector rotated by angle p (the greek letter that looks like an Owith a vertical bar in it).in vector notation. After expanding ts becomes r' = r*cos(p) + n*(n dot r)*(1-cos(p)) + (r cross n)*sin(p)as is indicated in the picture, where r=OP, n is the unit vectorequivalent of ON, r' is OQ, dot means the dot product and cross meansthe cross product.Ts is a perfectly valid working rotation formula in itself. But youasked for a matrix. Ts formula can be expanded to a matrix formlike ts:1- the first term, r*cos(p), in matrix notation becomes Id*cos(p)where Id is the 3x3 identity matrix.2- the second term, we can recognize it as the orthogonal projectionof vector r onto the unit vector n. If n=(a,b,c) then the matrix ofts orthogonal projection is written as [ a^2 a*b a*c ] [ a*b b^2 b*c ] [ a*c b*c c^2 ]and you can see by multiplying ts matrix by vector r on theright-hand side that the result will be the same as calculating n*(ndot r) in the second term. Of course you have to multiply ts matrixby (1-cos(p)) to get the full second term.(In a linear algebra course you might learn that all orthogonalprojection operators in R^3 are symmetric, and ts matrix issymmetric as expected).3- the trd term is the cross-product operator of r and n. Thematrix form is:[ 0 -c b ][ c 0 -a ][ -b a 0 ]where here again n=(a,b,c) the unit vector on the axis. You canverify by multiplying with r on the right-hand side that ts isequivalent to taking the crossproduct of r and n. Multiply bysin(p) to get the full trd term.(Ts matrix is anti-symmetric and here again in a linear algebracourse you might learn that all anti-symmetric operators in R^3 arecross-product operators).So if you add all the matrices in 1,2 and 3 you get:[ t*a^2+d t*a*b-s*c t*a*c + s*b ][ t*a*b + s*c t*b^2+d t*b*c - a*s ][ t*a*c - s*b t*b*c+s*a t*c^2 + d ]where t=(1-cos(p)), n=(a,b,c) the unit vector along the axis, s=sin(p),d=cos(p)p is the angle.Ts matrix has all the properties of a rotation:the determinant is 1the column vectors are orthonormal.the trace is (1+2*cos(p)) wch is a quick way to extract the angle.the eigenvalues are either 1,1 and 1, or 1,-1 and -1, or 1, exp(i*p)and exp(-i*p) depending on the value of p, so their product is 1(the determinant). === Subject: Re: matrices of revolution by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2FFefc00970;>i was trying to figure the dard matrix for rotating a vector>around another vector some number of radians in 3-D and i cant find>any where in my linear algebra a way to do it. ive found a few>different little 'nts' on how to do it i tnk. like if u find e1e2>and e3 (basically i j and k) and find what they become after the>rotation then the dard matrix will be [a1 a2 a3] where a1 a2 and>a3 are the new transformed vectors. but i have no idea how to find a1>a2 and a3. also, ive found how to do it around an axis. but im still>very confused on how to do it around any random vector.>any help would be nicelet the axis of rotation be given by the vector a=(a1,a2,a3) witha1^2 + a2^2 + a3^2 = 1.Let us first assume that the axis goes through (0,0,0).Let D1 be the rotation in the y-z-plane that transforms (a1,a2,a3)to (a1,sqrt(1-a1^2),0), D2 the rotation in the x-y-plane thattransforms (a1,sqrt(1-a1^2),0) to (1,0,0).Then the corresponding matrices D1 and D2 are given byD1(1,1) = 1, D1(2,2) = a2/sqrt(1-a1^2), D(2,3) = a3/sqrt(1-a1^2),D1(3,2) = -D1(2,3), D1(3,3) = D1(2,2), D(I,J) = 0 else,D2(1,1) = a1, D2(1,2) = sqrt(1-a1^2), D2(2,1) = -D2(1,2),D2(2,2) = D2(1,1), D2(3,3) = 1, D2(I,J) = 0 else.Then the matrix D for the rotation around (a1,a2,a3) with angle alphais given byD = (D2*D1)' * A * (D2*D1)where A is the usual rotation matrix for rotation in the y-z-planewith angle alpha:(A(1,1) = 1, A(2,2) = cos(alpha), A(2,3) = -sin(alpha), A(3,2) = -A(2,3), A(3,3) = A(2,2), A(I,J) = 0 else). Here ' means 'transpose'.Multiplying out you arrive atD(1,1) = a1^2 + (a2^2+a3^2)*cos(alpha)D(1,2) = -a3*sin(alpha) + a1*a2+(1-cos(alpha))D(1,3) = a2*sin(alpha) + a1*a3*(1-cos(alpha))D(2,1) = a3*sin(alpha) + a1*a2*(1-cos(alpha))D(2,2) = a2^2 + (a1^2+a3^2)*cos(alpha)D(2,3) = -a1*sin(alpha) + a2*a3*(1-cos(alpha))D(3,1) = -a2*sin(alpha) + a1*a3*(1-cos(alpha))D(3,2) = a1*sin(alpha) + a2*a3*(1-cos(alpha))D(3,3) = a3^2 + (a1^2+a2^2)*cos(alpha).If the axis of rotation does not go through the origin, choose a point x0 on the straight line spanned by the axis.Then the affine transform t(x) = x0 + D*(x-x0) with D from aboveis the rotation around a = (a1,a2,a3). === Subject: Re: matrices of revolution> i was trying to figure the dard matrix for rotating a vector> around another vector some number of radians in 3-D and i cant find> any where in my linear algebra a way to do it. ive found a few> different little 'nts' on how to do it i tnk. like if u find e1 e2> and e3 (basically i j and k) and find what they become after the> rotation then the dard matrix will be [a1 a2 a3] where a1 a2 and> a3 are the new transformed vectors. but i have no idea how to find a1> a2 and a3. also, ive found how to do it around an axis. but im still> very confused on how to do it around any random vector.> any help would be niceThe really hard way to do ts is directly with vectors and matrices. The really easy way to do ts is to first learn about quaternions, then represent your rotation as a quaternion, then convert that quaternion to a transform matrix. Check out the Matrix and Quaternion FAQ at. === Subject: Re: matrices of revolutioni was trying to figure the dard matrix for rotating a vector> around another vector some number of radians in 3-D and i cant find> any where in my linear algebra a way to do it. ive found a few> different little 'nts' on how to do it i tnk. like if u find e1 e2> and e3 (basically i j and k) and find what they become after the> rotation then the dard matrix will be [a1 a2 a3] where a1 a2 and> a3 are the new transformed vectors. but i have no idea how to find a1> a2 and a3. also, ive found how to do it around an axis. but im still> very confused on how to do it around any random vector.> any help would be niceThe really hard way to do ts is directly with vectors and matrices. The > really > easy way to do ts is to first learn about quaternions, then represent your > rotation as a quaternion, then convert that quaternion to a transform matrix. > Check out the Matrix and Quaternion FAQ at> .Not at all. For general rotatins quaternions are useful, but ts can easily be done with vectors. See, for example, 's post. === Subject: Re: matrices of revolution> Not at all. For general rotatins quaternions are useful, but ts can > easily be done with vectors. See, for example, 's post.The OP said rotate a vector around another vector some number of radians. How is that not a general rotation?-- If ts message helped you, consider buying an itemfrom my wish list: === Subject: Re: matrices of revolutionNot at all. For general rotatins quaternions are useful, but ts can > easily be done with vectors. See, for example, 's post.The OP said rotate a vector around another vector some number of radians. > How > is that not a general rotation?General rotations are often given as a sequence of separate rotations about separate axes of rotation, or in other similarly complex ways, rather that as a single rotation of a single vector about a single fixed axis. === Subject: Re: matrices of revolution> i was trying to figure the dard matrix for rotating a vector> around another vector some number of radians in 3-D and i cant find> any where in my linear algebra a way to do it. ive found a few> different little 'nts' on how to do it i tnk. like if u find e1 e2> and e3 (basically i j and k) and find what they become after the> rotation then the dard matrix will be [a1 a2 a3] where a1 a2 and> a3 are the new transformed vectors. but i have no idea how to find a1> a2 and a3. also, ive found how to do it around an axis. but im still> very confused on how to do it around any random vector.> any help would be niceTo rotate A around B: (1) project A onto B, getting C =B(A.B)/|B|^2, wch is parallel to B and whose length isproportional to |A| and independent of |B|; (2) take the crossproduct D = BxA/|B|, wch is perpendicular to both A and B (andthus C), and whose length is again proportional to |A| andindependent of |B|; (3) the rotated A' = C + (A - C).cos(theta) +D.sin(theta). (Note that the three terms on the RHS all havelengths proportional to |A| and independent of |B|, and aremutually perpendicular.) Ts can be easily turned into a matrixequation. === Subject: Re: matrices of revolutionok. for example:what is the dard matrix of revolution for revolving a vectoraround the vector <1,2,3> or some # of radians.I tnk you may have some difficulty getting an answer to ts -> certainly one from me. Can you say _precisely_ what you mean by> rotating a vector around another vector? === Subject: Re: matrices of revolutio> ok. for example:> what is the dard matrix of revolution for revolving a vector> around the vector <1,2,3> or some # of radians.I tnk you may have some difficulty getting an answer to ts -> certainly one from me. Can you say _precisely_ what you mean by> rotating a vector around another vector?To rotate one vector, A, about another,B, resolve A into a component C parallel to B and a component, D, perpendicular, to B.The rotated vector D will be a linear combination of D and (D cross B), say E, of the same length as D, but C will be unchanged, and the rotated vector will then be the sum C + E === Subject: AIR by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2FMT7t17960;What is 2xy+365+xy+zy-z=? (simplified) === Subject: AIR by support1.mathforum.org (8.11.6/8.11.6/The Math Forum, $Revision: 1.9 primary) id i2FMT7817964;Ts is an open chat room for Academy At Ivy Ridge students only. Formath questions only, not for chatting but for getting answeres.