Dunkum Posted December 7, 2015 Report Share Posted December 7, 2015 There's a lot more matrix theory here than I expected. Did you know that in my graduate linear algebra class, we used matrices maybe... twice? Turns out that you don't need (or want) matrices for proving powerful results.makes sense, honestly. they are an enormous pain to work with even when you just have integers in them. with variables, you end up with a ponderous number of ridiculously long expressions of multiple variables to manipulate. just teh basic algebra parts take forever, before you even bring in the theory and proof parts 1 Quote Link to comment Share on other sites More sharing options...
Chaos Posted December 9, 2015 Report Share Posted December 9, 2015 makes sense, honestly. they are an enormous pain to work with even when you just have integers in them. with variables, you end up with a ponderous number of ridiculously long expressions of multiple variables to manipulate. just teh basic algebra parts take forever, before you even bring in the theory and proof parts Linear transformations are where its at. Makes eigenvalue proofs much easier; no determinants required. 0 Quote Link to comment Share on other sites More sharing options...
Hood Posted December 11, 2015 Report Share Posted December 11, 2015 I really hate determinants now. I wish I had read, linear algebra done right. The author rarely uses determinants in it. 0 Quote Link to comment Share on other sites More sharing options...
Stormgate Posted December 12, 2015 Report Share Posted December 12, 2015 I'm in statistics right now... I got tired of entering information into lists, so I wrote a calculator program to do it for me. 2 Quote Link to comment Share on other sites More sharing options...
Chaos Posted December 16, 2015 Report Share Posted December 16, 2015 I really hate determinants now. I wish I had read, linear algebra done right. The author rarely uses determinants in it. "Linear Algebra Done Right" was in fact the exact book my grad class had used. It is a nice book. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted January 12, 2016 Author Report Share Posted January 12, 2016 A while back, I made a post on the cubic formula. Well, I'm proud to say that I've successfully managed to derive the quartic formula as well. What I'm going to due is list the identities that lead to the eventual calculation of x, then list the actual derivation below in a spoiler box. I wouldn't mind having someone checking my math to make sure that I'm right. ax^4+bx^3+cx^2+dx+e=0 p=c/a -3b^2/8a^2 q=b^3/8a^3 -bcy/2a^2 +d/a r=cb^2/16a^3 -3b^4/256a^4 +e/a -bd/4a^2 m=-p^2/12-1 n=pr/3 -p^3/107 -q^2/8 o=(-n/2+-(n^2/4+m^3/27)^1/2)^1/3 L=o-m/3o=(((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3)-((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2)))/12m^2 z=(((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3)-((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2))-10pm^2)/12m^2 s=(p+2(((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3) -((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2))-10pm^2)/12m^2)^1/2 t=((((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3)- ((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2))-10pm^2)/12m^2)^2 +2p(((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3) -((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2))-10pm^2)/12m^2+p^2-r)^1/2 y=(s+-(s^2-4(p-t+z)^1/2)/2 x=y-b/4a x^4+b/ax^3+c/ax^2+dx/a+e/a Let x = y=b/4a Expand: y^4 -by^3/a +3b^2y^2/8a^2 -4b^3y/64a^3 +b^4/256a^4 +by^3/a -6b^2y^2/8a^2 +12b^3y/64a^3 -4b^4/256a^4 +cy^2/a -bcy/2a^2 +cb^2/16a^3 +dy/a -bd/4a^2 +e/a Collect: y^4 +(c/a -3b^2/8a^2)y^2 +(b^3/8a^3 -bcy/2a^2 +d/a)y +(cb^2/16a^3 -3b^4/256a^4 +e/a -bd/4a^2) p q r Simplify: y^4 +py^2 +yqy +r=0 Complete square: (y^2+p)^2=py^2-qy+(p^2-r) Set discriminant to o: +(y^2+p+z)^2 +(y^2+p+z)^2 -(y^2+p)^2 -(y^2+p)^2 (y^2+p+z)^2=py^2+2zy^2-qy+z^2+2pz+p^2-r Discriminant=q^2-4(p+2z)(z^2+2pz+p^2-r) -8z^3-20pz^2-16p^2z+8rz-4p^3+4pr+q^2 z^3+5pz^2/2+2p^2z-rz+p^3/2-pr/2-q^2/8 let z = L-5p/6 Expand: L^3 -5pL^2/2 +25p^2L/12 -125p^3/216 +5pl/2 -50p^2L/12 +375p^3/216 +24p^2L/12-360p^3/216 -12rL/12 +5pr/6 +p^3/2 -pr/2 -q^2/8 Collect: L^3 +(-p^2/12-1)L +(pr/3-p^3/108-q^2/8) m n Simplify: L^3+mL +n Let L=o-m/30 Expand: o^3+n-m^3/27o^3 Simplify: (o^3)^2+no^3+m^3/27o^3 o=(-n/2+-(n^2/4+m^3/27)^1/2)^1/3 L=((-n/2+-(n^2/4+m^3/27)^1/2)^1/3)-m/(3(-n/2+-(n^2/4+m^3/27)^1/2)^1/3) Rationalize: ((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3)(-9n-(12m^3+81q^2)^1/2))/12m^2 L=(((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3)-((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2)))/12m^2 z=(((-216nm^2+24m^2(12m^2+81n^2)^1/2)^1/3)-((8m^3+54n^2-6n(12m^3+81n^2)^1/2)^1/3(-9n-(12m^3+81n^2)^1/2))-10pm^2)/12m^2 (y^2+p+z)^2=(sy+t)^2 y^2-sy=t-p-z y=(s+-(s^2-4(p-t+z)^1/2)/2 x=y/b/3a 1 Quote Link to comment Share on other sites More sharing options...
Dunkum Posted January 12, 2016 Report Share Posted January 12, 2016 ok, now see if you can do the same for the quintic 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted January 12, 2016 Author Report Share Posted January 12, 2016 ok, now see if you can do the same for the quintic Using resultants, you can reduce it down to the form x+a1x+a. From there, you can solve using elliptic functions. 0 Quote Link to comment Share on other sites More sharing options...
Dunkum Posted January 12, 2016 Report Share Posted January 12, 2016 Using resultants, you can reduce it down to the form x+a1x+a. From there, you can solve using elliptic functions. never seen that method. I have seen a proof that you can't get it down to just radicals, though. my abstract algebra textbook included it in one of the later chapters. well more like it took 5 of the later chapters to set it up and then prove it. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted January 12, 2016 Author Report Share Posted January 12, 2016 never seen that method. I have seen a proof that you can't get it down to just radicals, though. my abstract algebra textbook included it in one of the later chapters. well more like it took 5 of the later chapters to set it up and then prove it. Check these out http://math.stackexchange.com/questions/542108/how-to-transform-a-general-higher-degree-five-or-higher-equation-to-normal-form http://math.stackexchange.com/questions/540964/how-to-solve-fifth-degree-equations-by-elliptic-functions 0 Quote Link to comment Share on other sites More sharing options...
Kestrel Posted January 12, 2016 Report Share Posted January 12, 2016 (edited) When is simplifying sin/cos/tan equations relevant to anything. Why is pre-cal so weird and why does it teach concepts that I will never use out of high school. I'm not even sure what I don't understand about it, I just don't understand this unit. But physics has gotten easier. Edited January 12, 2016 by LarkoftheRiver 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted January 12, 2016 Author Report Share Posted January 12, 2016 When is simplifying sin/cos/tan equations relevant to anything. Why is pre-cal so weird and why does it teach concepts that I will never use out of high school. I'm not even sure what I don't understand about it, I just don't understand this unit. But physics has gotten easier. Which unit? 0 Quote Link to comment Share on other sites More sharing options...
Kestrel Posted January 12, 2016 Report Share Posted January 12, 2016 In physics? Its work right now. Pre-cal is just unit mess. 0 Quote Link to comment Share on other sites More sharing options...
Dunkum Posted January 12, 2016 Report Share Posted January 12, 2016 When is simplifying sin/cos/tan equations relevant to anything. Why is pre-cal so weird and why does it teach concepts that I will never use out of high school. I'm not even sure what I don't understand about it, I just don't understand this unit. But physics has gotten easier. most math after algebra 1 or geometry is really more about preparing future science, math, and engineering students for the stuff they will need. it honestly doesn't have much direct day to day relevance to anyone not in a field related to one of those (though if you ever end up doing your own home improvement work, you may find yourself calling upon a little bit of trigonometry as well. it has a lot of relevance for anything involving triangles). that said, I personally think a lot of it is interesting. limits and derivatives and power series can provide some bizarre results, like Euler's identity. and they also serve as a basis for other, more interesting concepts. but then, I majored in Math. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted January 12, 2016 Author Report Share Posted January 12, 2016 In physics? Its work right now. Pre-cal is just unit mess. My class just started on that as well. What are you having trouble understanding? 0 Quote Link to comment Share on other sites More sharing options...
Oversleep Posted February 12, 2016 Report Share Posted February 12, 2016 (edited) I recently have thought how certain Shards have numbers associated with them and about the concept of Shards as four dimensional vectors (axis of change (Cultivation, Ruin) vs. no change (Preservation) and similar).What if these numbers are just Shard's coordinates? Honor is 0;1;0;1;0 (ten in binary); Preservation 1;0;0;0;0; (sixteen in binary). I am not sure about Endowment, but maybe five is her number? Then Endowment is 0;0;1;0;1. Maybe combining Shards is a logical operation? What if all the Shards together (Adonalsium) have 0;0;0;0;0 (or 1;0;0;0;0)? 1 xor 2 xor 3 xor... xor 15 xor 16 is 16 (Shards' numbers are 1 to 16 and Adonalsium is 16) 0 xor 1 xor 2 xor... xor 14 xor 15 is 0 (Adonalsium is 0 and Shards are 0 to 15 and Shard's number is number+1)Maybe it's not xor. I don't know, random idea.Maybe it's a different binary code (there are many), since Preservation would be the only one with a 1 in 2^4 slot in the natural one (with Shards being numbered 1-16. On the other hand, with Shard being numbered 0-15 Preservation would be 0;1;1;1;1 - so many 1's).What are your thoughts? Edited February 13, 2016 by Oversleep 1 Quote Link to comment Share on other sites More sharing options...
Aonar Posted April 1, 2016 Report Share Posted April 1, 2016 (edited) Alright. So in physics class the other day we got an assignment; basically, we have to build a novel/useful electronic device based on the motor principle. Naturally, my mind went to a desktop railgun. Now, here's the problem. I have access to a handful of 1 mF, 25V capacitors for free. Assuming a resistance in the railgun of about 0.01 Ohms (I haven't actually built it yet) that should give me about 2500A to work with per capacitor; more than enough for my purposes, even given nothing like ideal conditions and a high probability of error. However, I'll have 2500A for about 10 microseconds, tops. Now, assuming I've done my math right in my estimations, can anyone tell me if 2500A (or 5 Newtons of force, after all the calculations are done) for 10 microseconds is enough to do anything meaningful?(For reference, I'm assuming a projectile weight between 10 and 100 grams) My gut instinct says no, but I want to be sure before I start ordering higher-quality capacitors. Edited April 1, 2016 by Aonar Faileas 0 Quote Link to comment Share on other sites More sharing options...
Dunkum Posted April 1, 2016 Report Share Posted April 1, 2016 (edited) Alright. So in physics class the other day we got an assignment; basically, we have to build a novel/useful electronic device based on the motor principle. Naturally, my mind went to a desktop railgun. Now, here's the problem. I have access to a handful of 1 mF, 25V capacitors for free. Assuming a resistance in the railgun of about 0.01 Ohms (I haven't actually built it yet) that should give me about 2500A to work with per capacitor; more than enough for my purposes, even given nothing like ideal conditions and a high probability of error. However, I'll have 2500A for about 10 microseconds, tops. Now, assuming I've done my math right in my estimations, can anyone tell me if 2500A (or 5 Newtons of force, after all the calculations are done) for 10 microseconds is enough to do anything meaningful?(For reference, I'm assuming a projectile weight between 10 and 100 grams) My gut instinct says no, but I want to be sure before I start ordering higher-quality capacitors. well, I mean it's been a while since I did physics, and my gen physics class never got into the whole electricity stuff, that was saved for more advanced classes that I didn't take, but I do know F=ma. seems like, even given a 10 g object, 5 newtons should give 500 m/s^2 of acceleration, which starting from rest and lasting 10 microseconds would get you to a grand total final speed of something like 5 millimeters per second, I think. unless I am missing something Edited April 1, 2016 by Dunkum 0 Quote Link to comment Share on other sites More sharing options...
Aonar Posted April 2, 2016 Report Share Posted April 2, 2016 (edited) well, I mean it's been a while since I did physics, and my gen physics class never got into the whole electricity stuff, that was saved for more advanced classes that I didn't take, but I do know F=ma. seems like, even given a 10 g object, 5 newtons should give 500 m/s^2 of acceleration, which starting from rest and lasting 10 microseconds would get you to a grand total final speed of something like 5 millimeters per second, I think. unless I am missing something Okay, thanks. So I know both capacitance and amperage should increase velocity quadratically. (Although capacitance is limited by the amount of time the projectile contacts the rails.) So if I add another 10 capacitors in parallel, I should get a speed 100x higher. That's about 1/2m/s. If I do 20 I should get about 1m/s, or 3.6 km/h. That's still not much, but it's at least in numbers that can be easily visualized. If I'm willing to shell out the money for better capacitors (probably 25V 1 cF, if I can find some to that specification) I could easily get some pretty decent speeds, even with heavier projectiles. Edited April 2, 2016 by Aonar Faileas 0 Quote Link to comment Share on other sites More sharing options...
Oversleep Posted April 2, 2016 Report Share Posted April 2, 2016 (edited) There's a serious lack of mad-scientist-laughter. C'mon!>Get assignment at physics class to build something>First though : Railgun! AHAHAHAHAHAHA!!!!!! Edited April 2, 2016 by Oversleep 2 Quote Link to comment Share on other sites More sharing options...
Dunkum Posted April 2, 2016 Report Share Posted April 2, 2016 Okay, thanks. So I know both capacitance and amperage should increase velocity quadratically. (Although capacitance is limited by the amount of time the projectile contacts the rails.) So if I add another 10 capacitors in parallel, I should get a speed 100x higher. That's about 1/2m/s. If I do 20 I should get about 1m/s, or 3.6 km/h. That's still not much, but it's at least in numbers that can be easily visualized. If I'm willing to shell out the money for better capacitors (probably 25V 1 cF, if I can find some to that specification) I could easily get some pretty decent speeds, even with heavier projectiles. if capacitors increase velocity quadratically, then 20 should get you 400x the rate, which would be 2 m/s or a fast walking pace. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted April 8, 2016 Author Report Share Posted April 8, 2016 TOday, I learned about the formula sin(a/2)=(1/2*1-cos a)^1/2. My question is: how could one tweak it for sin(a/b)? 0 Quote Link to comment Share on other sites More sharing options...
Chaos Posted April 8, 2016 Report Share Posted April 8, 2016 TOday, I learned about the formula sin(a/2)=(1/2*1-cos a)^1/2. My question is: how could one tweak it for sin(a/b)? Irreparably, cannot be tweaked in that sense. That formula has a very specific derivation that requires a/2. 0 Quote Link to comment Share on other sites More sharing options...
Silverblade5 Posted April 8, 2016 Author Report Share Posted April 8, 2016 Irreparably, cannot be tweaked in that sense. That formula has a very specific derivation that requires a/2. In that case, do you know how the formula was derived? 0 Quote Link to comment Share on other sites More sharing options...
Haelbarde Posted April 8, 2016 Report Share Posted April 8, 2016 In that case, do you know how the formula was derived? http://www.themathpage.com/atrig/double-proof.htm If you were to attempt it, you'd have to apply the sum indentity b times, and then see what you got. Probably not going to be super neat... 0 Quote Link to comment Share on other sites More sharing options...
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