/a/ - Anime & Manga » Thread #132918120

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Anonymous Fri 30 Oct 2015 12:48:21 No.132918120 View Reply Report Delete

Quoted By: >>132918163 >>132918206 >>132918254 >>132918384 >>132920247 >>132920787

Easy maths /a/, only 1st year ou 2nd year of college !

derp_commander !!SSXdvR3LCoa

derp_commander !!SSXdvR3LCoa Fri 30 Oct 2015 12:49:46 No.132918163 Report Delete

Quoted By: >>132918202

>>132918120
in what major?

Anonymous

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Anonymous Fri 30 Oct 2015 12:51:06 No.132918202 Report Delete

Quoted By: >>132920596

>>132918163

Mhmm m/a/ths, physics and computer science I guess.

Anonymous

Anonymous Fri 30 Oct 2015 12:51:09 No.132918206 Report Delete

Quoted By: >>132918248

>>132918120
This problem is sloppily formulated and the grammar is shit. I'm not even sure where to start.

Anonymous

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Anonymous Fri 30 Oct 2015 12:52:27 No.132918241 Report Delete

Quoted By:

>differential of det
What??

derp_commander !!SSXdvR3LCoa

derp_commander !!SSXdvR3LCoa Fri 30 Oct 2015 12:52:37 No.132918248 Report Delete

Quoted By:

>>132918206
It's a simple expression of the extent of the commutativity of matrix addition.

Anonymous

Anonymous Fri 30 Oct 2015 12:52:45 No.132918254 Report Delete

Quoted By:

>>132918120
I'm fucked next semester. I forgot everything about math.

Anonymous

Anonymous Fri 30 Oct 2015 12:55:29 No.132918348 Report Delete

Quoted By: >>132918402 >>132918583

det A means the determinant of the matrix A of course !

Anonymous

Anonymous Fri 30 Oct 2015 12:56:37 No.132918384 Report Delete

Quoted By: >>132918420

>>132918120
get someone else to do your homework

Anonymous

Anonymous Fri 30 Oct 2015 12:57:10 No.132918402 Report Delete

Quoted By:

>>132918348
What a genius you are.

Anonymous

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Anonymous Fri 30 Oct 2015 12:57:44 No.132918420 Report Delete

Quoted By: >>132919031

>>132918384

I already did them.

Anonymous

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Anonymous Fri 30 Oct 2015 13:02:29 No.132918583 Report Delete

Quoted By: >>132918699

>>132918348
Oh, I found how I can do it.

Anonymous

Anonymous Fri 30 Oct 2015 13:05:47 No.132918699 Report Delete

Quoted By:

>>132918583

You mean finding the differential of det ?

Well, there are two questions in this problem, but I think that both are linked.

But it is not the differential of det. Because a differential is a linear application.

Anonymous

Anonymous Fri 30 Oct 2015 13:12:21 No.132918910 Report Delete

Quoted By:

Please /a/ ! ;_;

Anonymous

Anonymous Fri 30 Oct 2015 13:16:32 No.132919031 Report Delete

Quoted By: >>132919073

>>132918420
Let M be the matrix ((0,a,0),(-a,0,b),(0-b,0)).

Now your problem is just x' = Mx. Which you really should be able to solve.

Anonymous

Anonymous Fri 30 Oct 2015 13:18:00 No.132919073 Report Delete

Quoted By: >>132919193

>>132919031

Indeed, it is the good way. Great Andrew !

Now you must calculate exp M. :3

Anonymous

Anonymous Fri 30 Oct 2015 13:18:25 No.132919081 Report Delete

Quoted By:

shit's so useless

Anonymous

Anonymous Fri 30 Oct 2015 13:21:43 No.132919193 Report Delete

Quoted By: >>132919331

>>132919073
>Now you must calculate exp M
Nah, problem's solved. What do you think I am, an engineer?

Anonymous

Anonymous Fri 30 Oct 2015 13:25:53 No.132919331 Report Delete

Quoted By:

>>132919193

But is M is not diagonalizable it is rather hard. Because trigonalization is a much more harder transformation than diagonalization.

And what about the two other problems Andrew-san ?

PS : You are in a computer science department ?

Anonymous

Anonymous Fri 30 Oct 2015 13:27:41 No.132919406 Report Delete

Quoted By: >>132919453

Does /a/stronomy count?

Anonymous

Anonymous Fri 30 Oct 2015 13:28:56 No.132919453 Report Delete

Quoted By:

>>132919406

Yes, of course. But /a/stronomy with physics and m/a/ths.

Anonymous

Anonymous Fri 30 Oct 2015 13:31:26 No.132919535 Report Delete

Quoted By: >>132919661

At least the past few m/a/th threads started with screenshots of bad anime equations on blackboards, THIS is just literally putting college textbook questions onto the same picture over and over again with MSPaint.

MAKE IT AT LEAST A TINY LITTLE BIT /a/ WOULD YOU

Anonymous

Anonymous Fri 30 Oct 2015 13:35:24 No.132919661 Report Delete

Quoted By: >>132919723

>>132919535
>literally putting college textbook questions onto the same picture
Not even that, no textbook would be that poorly phrased.

Anonymous

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Anonymous Fri 30 Oct 2015 13:37:13 No.132919723 Report Delete

Quoted By:

>>132919661

Poorly phrased ?

Anonymous

Anonymous Fri 30 Oct 2015 13:52:19 No.132920247 Report Delete

Quoted By: >>132920508

>>132918120
I'm gonna use X = aI for any a in the field and show that A and B have the same eigenvalues. Now in an algebraically closed field this means that the minimal polynomials of A and B coincide by Cayley-Hamilton, therefore they leave eigenspaces invariant. Now let X be projections onto any of these eigenspaces, and if there exists a vector v such that Av =! Bv, then A + X =! B + X such that their determinants would not equal, which is a contradiction. Thus A and B correspond to the same linear transformation T: V -> V.

Anonymous

Anonymous Fri 30 Oct 2015 13:57:58 No.132920431 Report Delete

Quoted By:

Man, I kinda feel bad that I suck at math.

Anonymous

Anonymous Fri 30 Oct 2015 14:00:24 No.132920508 Report Delete

Quoted By: >>132920642

>>132920247
I think this works, but it's been about 8 years since I last did any proper linear algebra.

Do you mind explaining why Av =! Bv would imply the determinants differ? I'm probably just being thick...

Anonymous

Anonymous Fri 30 Oct 2015 14:03:20 No.132920596 Report Delete

Quoted By: >>132920734

>>132918202
I'm going to use Cauchy-Schwartz inequality on f*f", such that |f*f"|^2 <= |f| *|f"|. Now use integration by parts to massage the left member into 2*(|(f'+f)*f'|).

Anonymous

Anonymous Fri 30 Oct 2015 14:05:16 No.132920642 Report Delete

Quoted By:

>>132920508
This is only the case if they leave the same spaces invariant. If they differ then the eigenpolynomials would have different linear factos

Anonymous

Anonymous Fri 30 Oct 2015 14:08:08 No.132920734 Report Delete

Quoted By: >>132920853

>>132920596

Good idea. You can prove f' is Riemann-integrable with an integration by parts. Of course, for the inequality, it is Cauchy-Schwartz.

Anonymous

Anonymous Fri 30 Oct 2015 14:10:01 No.132920787 Report Delete

Quoted By:

>>132918120
This one isn't even hard, I just can't remember the formula for the determinant.

You just calculate the determinant with some numbers a,b,c and show that you'll get the same answer in both sides.

Anonymous

Anonymous Fri 30 Oct 2015 14:12:03 No.132920853 Report Delete

Quoted By:

>>132920734
If f" in C implies f' is in C^1, so f is square integrable (summable, even)

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