Jump to content

Math and science


Silverblade5

Recommended Posts

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
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

  • 4 weeks later...

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

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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

Link to comment
Share on other sites

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 by LarkoftheRiver
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

  • 5 weeks later...

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 by Oversleep
Link to comment
Share on other sites

  • 1 month later...

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. :P

 

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 by Aonar Faileas
Link to comment
Share on other sites

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. :P

 

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 by Dunkum
Link to comment
Share on other sites

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 by Aonar Faileas
Link to comment
Share on other sites

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.

Link to comment
Share on other sites

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.

Link to comment
Share on other sites

Join the conversation

You can post now and register later. If you have an account, sign in now to post with your account.

Guest
Reply to this topic...

×   Pasted as rich text.   Paste as plain text instead

  Only 75 emoji are allowed.

×   Your link has been automatically embedded.   Display as a link instead

×   Your previous content has been restored.   Clear editor

×   You cannot paste images directly. Upload or insert images from URL.

Loading...
  • Recently Browsing   0 members

    • No registered users viewing this page.
×
×
  • Create New...