/sci/ - Science & Math

No

I'm high out of my mind so I can't string together a coherent explanation, however, the answer is quite obviously no. Just look at the Leibniz expression for a determinant and it's quite obvious.

>>4293667
>>4293692

Now taking bets on how many people will claim to be able to solve the problem before someone really does.

>>4293664
>\left[ \displaystyle{\vcenter{\rlap{\strut \ulap{\qquad\atop\smash{ 3 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ 1 }} }\rlap{\lower{1.5em}{\ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 4 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ 1 }} }}\rlap{\lower{3em}{\ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 5 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ 1 }} }}\rlap{\lower{4.5em}{\ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 6 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ 1 }} }}\rlap{\lower{6em}{\ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \ddots }} \ulap{\qquad\atop\smash{ \vdots }} }}\lower{7.5em}{\strut \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ n+1 }} }}}\right].

Thanks for the crippled, borderline-useless latex feature, moot.

Subtract the last row from each of the others:
<div class="math">\left[ \displaystyle{\vcenter{\rlap{\strut \ulap{\qquad\atop\smash{ 2 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }\rlap{\lower{1.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 3 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{3em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 4 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{4.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 5 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{6em}{\ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \ddots }} \ulap{\qquad\atop\smash{ \vdots }} }}\lower{7.5em}{\strut \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ n+1 }} }}}\right]</div>
(continued...)

Now subtract <span class="math">1/2[/spoiler] times the first row from the last row:
<div class="math">\left[ \displaystyle{\vcenter{\rlap{\strut \ulap{\qquad\atop\smash{ 2 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }\rlap{\lower{1.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 3 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{3em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 4 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{4.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 5 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{6em}{\ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \ddots }} \ulap{\qquad\atop\smash{ \vdots }} }}\lower{7.5em}{\strut \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ n+1+\frac{n}{2} }} }}}\right]</div>
(continued...)

Then subtract <div class="math">1/3</div> times the second row from the last row:
<div class="math">\left[ \displaystyle{\vcenter{\rlap{\strut \ulap{\qquad\atop\smash{ 2 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }\rlap{\lower{1.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 3 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{3em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 4 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{4.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 5 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{6em}{\ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \ddots }} \ulap{\qquad\atop\smash{ \vdots }} }}\lower{7.5em}{\strut \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ 1 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ n+1+\frac{n}{2}+\frac{n}{3} }} }}}\right]</div>
(continued...)

Continue, until you get:
<div class="math">\left[ \displaystyle{\vcenter{\rlap{\strut \ulap{\qquad\atop\smash{ 2 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }\rlap{\lower{1.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 3 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{3em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 4 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{4.5em}{\ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 5 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ -n }} }}\rlap{\lower{6em}{\ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \vdots }} \ulap{\qquad\atop\smash{ \ddots }} \ulap{\qquad\atop\smash{ \vdots }} }}\lower{7.5em}{\strut \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ 0 }} \ulap{\qquad\atop\smash{ \cdots }} \ulap{\qquad\atop\smash{ \sum_{m=1}^{n}\frac{n}{m} }} }}}\right]</div>
None of these operations has changed the value of the determinant, and since this is now a triangular matrix the determinant is just the product of the diagonal elements:
<div class="math">D_n = (n-1)! \sum_{m=1}^{n}\frac{n}{m} = n! \sum_{m=1}^{n}\frac{1}{m}</div>
Therefore <span class="math">D_n = \sum_{m=1}^{n}\frac{1}{m}[/spoiler] which is the harmonic series and is unbounded, of course. The answer is no!

After some obvious row reduction,
D_n = (2n+1)n!/2 + n!(-1)^(n-1)
= n! [ n + 1/2 + (-1)^n ]

So the answer is NO.

>>4293747
That last equation should be <div class="math">\frac{D_n}{n!} = \sum_{m=1}^{n}\frac{1}{m}</div>.

Testing LaTeX:

\frac{x}{y}

I'm guessing this won't work.

>>4293751
Read the fifth bullet point under the post form.

>>4293749
>D_n = n! [ n + 1/2 + (-1)^n ]

Hmm, you seem to have messed up your "obvious" row reductions...

See the five posts above yours written by someone who actually seems to know what he's talking about.

>>4293749
oops, I meant:
D_n = (2n+1)n!/2 + n!(-1)^(n-2)

The last 2 was a 1 above.

>>4293755
Wait, I have an error...hold on.

>>4293758
But you said it was obvious! How could you have effed up something so easy!

>>4293750
So the set <span class="math">\displaystyle{\left\{D_n \over n!\right\}_{n \geq 2}}[/spoiler] is not bounded (from both sides), but it is bounded from below because <span class="math">\frac{D_n}{n!} \geq \frac{3}{2}[/spoiler].

png transparency blacks out?

>>4293761
I meant obvious choices of operations. Saved myself a lot of time.

I don't understand why you have an attitude problem. This is a math problem solving forum.

>>4293765
I mean I saved time by not writing out the steps.

>>4293765
>I meant obvious choices of operations. Saved myself a lot of time.

Yeah, except your answer was completely wrong (except the "no" part) and you still haven't fixed it

>I don't understand why you have an attitude problem. This is a math problem solving forum.

Posting "it's obvious" isn't "solving" anything.

>>4293771
I was giving a brief outline. What is your problem?

I'm sorry I offended you by using the word "obvious". Perhaps I should have said "natural choices of row operations to simplify the matrix".

There was a mistake, in my original arithmetic, but I see how to correct it. It's slightly simpler with the correction:

D_n = (n-1)! [ 1 + n H_n ]

where H_n is the harmonic number.

Since H_n is unbounded, the answer is NO.

>>4293791
Now that I've solved it, I checked the other posted solution, which is essentially what I did. However, there is an error in
>>4293747
The (n-1),(n-1) entry should be 1 + (what's already there). This, of course, doesn't change the answer to the question.

>>4293791
>where H_n is the harmonic number.
Meant to say:
where H_n is the nth harmonic number. Obviously.

>>4293799
No, I absorbed the one into the sum by changing the upper index limit from <span class="math">n-1[/spoiler] to <span class="math">n[/spoiler]. Make sure you remember that the matrix is (n-1)x(n-1), not (n)x(n).
Let me clarify. That lower right element of the matrix is equal to:
<div class="math">n+1+\frac{n}{2} + \frac{n}{3} + \frac{n}{4} + \cdots + \frac{n}{n-1}</div>
<div class="math">=\frac{n}{1}+1+\frac{n}{2} + \frac{n}{3} + \frac{n}{4} + \cdots + \frac{n}{n-1}</div>
<div class="math">=\frac{n}{1}+\frac{n}{2} + \frac{n}{3} + \frac{n}{4} + \cdots + \frac{n}{n-1} + \frac{n}{n}</div>
<div class="math">=\sum_{m=1}^{n}\frac{n}{m}</div>

>>4293807
Ah yes, you are right! I did have an n/n term in mine that shouldn't have been there.

<span class="math">
D_n = (n-1)! \sum_{j=1}^n \frac{n}{j}
[\math][/spoiler]

Test:
<div class="math">
D_n = (n-1)! \sum_{j=1}^n \frac{n}{j}
<div class="math"></div></div>

Test:
<div class="math">
\begin{eqnarray}
D_n &= (n-1)! \sum_{j=1}^n \frac{n}{j}\\
&= n! H_n.
<div class="math"></div></div>

Test:
<div class="math">
\begin{eqnarray}
D_n &= (n-1)! \sum_{j=1}^n \frac{n}{j}\\
&= n! H_n.
\end{eqnarray}
<div class="math"></div></div>

Test:
<div class="math">
D_n &= (n-1)! \sum_{j=1}^n \frac{n}{j}
&= n! H_n.
<div class="math"></div></div>

Test:
<div class="math">
D_n = (n-1)! \sum_{j=1}^n \frac{n}{j}
= n! H_n.
<div class="math"></div></div>

>>4293829
>>4293826
>>4293825
>>4293823
>>4293821
>>4293816
hey, moran. click the jsmath button in the bottom right corner and 'test' in the 'interactive lab' (home page).

Damn, this one was the kind of problems I like, and I'm too late... Nice work, anon.
Also, link to the \matrix to jsmath converter?

>>4293926

>>4293971
I was talking about the tool that makes matrices from you from the "normal" LaTeX syntax into the awful jsMath syntax.

>>4293995
just use LaTeX syntax, it works in /sci/

>>4294019
No, it doesn't.
Examples:
Identity matrix of dimension (2,2) using an array:
\left(\begin{array}{cc}1&0\\0&1\end{array}\right)
<span class="math">\left(\begin{array}{cc}1&0\\0&1\end{array}\right)[/spoiler]

With \matrix:
\begin{matrix}1&0\\0&1\end{matrix}
<span class="math">\begin{matrix}1&0\\0&1\end{matrix}[/spoiler]

Doesn't work. Just look at how OP's matrix is written. Writing in this form is annoying. A tool exists that does the conversion. I'm looking for it.

why do you people say this is high school level math
this, at least, is linear algebra

>>4294212
It's not high school level, but it should be doable by a relatively good 1st year or 2nd year math student. Maybe with more time spent than for someone with more experience in that kind of problems, but it's the kind of exercise I would give to a relatively good student for a "40 min on the blackboard" test. And I would be disappointed if I had to give him a hint.

>>4294212
In most european countrys basic linear algebra is taught in high school.

>>4294225
Not in France. Well, I admit that high school math education is not really good in France, unlike university level math education, which is kinda good.

I can't really speak for other European countries, though. But I know Linear Algebra used to be taught in French high schools until a few decades ago (I'd say until the end of the 70ies maybe, I guess google or older French persons could confirm the exact date).

>>4293871
>moran

Moron for not divining the purpose of a tiny, hard-to-see button at the bottom right corner of the page. That makes sense.

>>4293871
Pic related, it's a moran.

>>4294265
Not in The Netherlands either, maybe it's taught in one of the Scandinavian countries?

>>4295448
Some very basic linear algebra was taught in Norwegian high schools 6 years ago, if the student specialized in math that is. Some recent reforms may have changed it though

>>4295534
I'm curious, what do you mean by basic linear algebra?

>>4295368
This is Moran.

>>4295557
>Vectors
>Lines and planes (parameter and coordinate representation)
>Positional relationships of points, lines and planes in space
>Straight lines, planes
>Scalar product with applications
>Length of a vector
>Angle between two vectors, orthogonality
>Vector product with applications
>Normal form of straight lines and plane equations
>Spacing provisions
>cutting angle
>Applications of linear systems
>Systematic solution methods, the structure and geometric interpretation of the solution set
>Concept of vector space
>Basis and dimension
>Term of the matrix product of matrices, inverse matrix, applications in geometry and non-geometric problems

This is what is currently taught in the second half of grade 11 in germany(google translated).

>>4295610
with introduction to complex numbers?

>>4295626
No, here is the link
http://www.kultusministerium.hessen.de/irj/servlet/prt/portal/prtroot/slimp.CMReader/HKM_15/HKM_Inte
rnet/med/491/491704b5-267f-121a-eb6d-f191921321b2,22222222-2222-2222-2222-222222222222,true

>>4293926
http://userscripts.org/scripts/show/111163

>>4295673
Cool, thanks!

>>4295610


This is how my education worked during high school:

9th) trig
10th) "precalc"
11th) "Calc 1" (basically just derivatives, ODEs and intro to integrals)
12th) "Calc 2" (integrals, series/sequences, very simple complex analysis, ODEs and PDEs)

Because my college did not accept my transfer credit (those 2 calc courses were technically supposed to be "city college classes" that were taught at my school), I still had to retake the basic calculus courses that you can normally skip with the AP calculus tests.

The college was a very good, but not top of the tier, UC (eg: not UCLA, UCB, or UCSD, but right below those in terms of ranking).

I did not consider that spending $280 ($140 for 2 AP tests) to take 2 tests was worth the time to slack off and take some retarded GE classes during my first 2 quarters of university.

311111411111511111611111n+1
Is the set n!Dnn2b

>>4296584
Totally.

Just in case anybody is still lost, I solved it another way.

If you subtract row 1 from all the others then cofactor expand the bottom row, your row reduction is a lot cleaner and you get that the magnitude of the determinant is strictly bounded from below by |(n+1)! - n!| which obviously diverges.

>>4295673
Thanks, but this application only works on Windows. [sigh]