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>> No.6342196 [View]
File: 119 KB, 903x986, riemann.jpg [View same] [iqdb] [saucenao] [google]
6342196

1. Euclid
2. Riemann
3. Cartan
4. Gauss

>> No.5102999 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann.jpg [View same] [iqdb] [saucenao] [google]
5102999

hello

>> No.4876692 [View]
File: 119 KB, 903x986, Riemann.jpg [View same] [iqdb] [saucenao] [google]
4876692

"Hiervon wäre allerdings ein strenger Beweis zu wünschen; ich habe indess die Aufsuchung desselben nach einegen flüchtigen vergeblichen Versuchen vorläufig bei Seite gelassen, da er für den nächsten Zweck meiner Untersuchung entbehrlich schien."

Is this the greatest cop-out of all time, or is it an acceptable admission of limitations?

>> No.4621977 [View]
File: 119 KB, 903x986, Lebesgue..jpg [View same] [iqdb] [saucenao] [google]
4621977

Sup guys.

I'm a graduate student in computer science.

One of my teachers said this, during type theory course:
There is no confluence in the Integral Calculus.

I can imagine why and I can wait until monday to ask my teacher directly.
But If someone already know... I'm listening.

>> No.4270902 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann..jpg [View same] [iqdb] [saucenao] [google]
4270902

<span class="math">\frac{\sqrt{1842603-690 \sqrt{5831385}}}{60}[/spoiler]

<span class="math">e^{ \pi } - \pi[/spoiler]

<span class="math">\pi^2 + \frac{\pi}{24}[/spoiler]

<span class="math">\sin(11)[/spoiler]

<span class="math">\cos(\ln( \pi + 20))[/spoiler]

<span class="math">2cos(cos(cos(cos(cos(cos(cos5))))))^{2}[/spoiler]

<span class="math">e^{6} - \pi^{4} - \pi^{5}[/spoiler]

<span class="math">\frac{\pi^{9}}{e^{8}}[/spoiler]

<span class="math">\pi e^{\frac{1}{\pi} \cdot \int_{0}^{1} \arctan{\frac{4 \pi t - 4 \pi t^{2} {\rm{artanh}} t}}{3\pi^{2} t^{2} + 4t^{2} {\rm{artanh}} t} - 8t {\rm{artanh}} t} +4}} \frac{dt}{t}}[/spoiler]

>> No.4157426 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann[1].jpg [View same] [iqdb] [saucenao] [google]
4157426

How does one find the correct stoichiometry of reactions without trial and error?

example:
Cr2O3 + Na2CO3 + KNO3 --> Na2Cr2O4 + CO2 + KNO2

>> No.4048909 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann[1]..jpg [View same] [iqdb] [saucenao] [google]
4048909

Which of these curves is the longest?

g1: [0,1] -> R, x -> 3x^2

g2: [0,1] -> R, x -> 2x^6

g3: [0,1] -> R, x -> x^4

Now, to calculate each individual length I would insert each funtion into:

L0,1=int( sqrt( 1+(g')^2 )) from 0 to 1

Trouble is, integrating the functions by hand is a mess and during a test I don't have the time nor the possibility to use a calculator or similar.
Is there a quicker way to find out which answer meats the criteria?

>> No.2332951 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann.jpg [View same] [iqdb] [saucenao] [google]
2332951

Can you believe some med student stole my fucking idea?

>> No.1643397 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann..jpg [View same] [iqdb] [saucenao] [google]
1643397

Ok, this has been bothering me for a while now. The Riemann integral is defined as:

<span class="math"> \int_{a}^{b}f(x)\mathrm{d}x=\lim_{\left \| P \right \|\rightarrow 0}\sum_{k=1}^{n}f(x_i^*)(x_i-x_{i-1}) [/spoiler]

Where <span class="math"> P=\left \{ a=x_0<\cdots<x_n=b \right \} [/spoiler], <span class="math"> x_i^*\in[x_{i-1},x_i] [/spoiler], and <span class="math"> \left \| P \right \| [/spoiler] is the mesh of the partition <span class="math"> P [/spoiler]. My question is, how do we use this to prove:

<span class="math"> \int_{a}^{a}f(x)\mathrm{d}x=0 [/spoiler]

I guess my real problem here is how you're supposed to define a partition of the interval <span class="math"> [a,a] [/spoiler]. I figured maybe it's defined as <span class="math"> P=\left \{ a,\cdots,a \right \} [/spoiler], but I can't seem to find anything about this online.

>> No.1441927 [View]
File: 119 KB, 903x986, Georg_Friedrich_Bernhard_Riemann.jpg [View same] [iqdb] [saucenao] [google]
1441927

Well I'll just leave this here...
http://arxiv.org/ftp/arxiv/papers/1006/1006.0381.pdf

>> No.1385058 [View]
File: 119 KB, 903x986, Riemann..jpg [View same] [iqdb] [saucenao] [google]
1385058

>>1385023

Yeah, we've met. My foot, meet your ass.

>> No.1283475 [View]
File: 119 KB, 903x986, Riemann..jpg [View same] [iqdb] [saucenao] [google]
1283475

LESS FILLING
MORE MEAT

0.60489864342163037024726591423595549975976254513024738037854664808218725349506035732740395691834955
4383033720410335824053109825223299163504834897013205000061739072732967872328610177617498177335444554
3606883969929973304849868201403341704424823003288609329808955898094142368486885480076564792446996920
5211820264021311385327273192475955245867887282863680644239664070439254271702144655807290763846702640
0849127468503001199073673321118407466675334069928433353541412283641533187892261583174774526078028289
1143767638766117462620783296161513144166327218391394308536435963214042571495394276189130443540597066
6197336861262266970275705911797430923034740128941752505546415294220502023526592860958269866161235317
6776315113423028511650527004252864766478852750645168901024356917560247452105590001096864598424902954
9066295487702415183425403771007703314842373064255400669267188421586454543284257956710219251830254463
0812910324491554672818928389496851399499316845536322840846966848594508659601467613962070363332940761
5668297062970071096240816727052753486858757850781202668122807183983469671446254821289164497797538709
9890823030491024586229073689415431140936786592885344091971079286873264986248035703327210118836985256
0431154037786592218858913530675928228389670848797218307879546572752260594712744701316776727200110536
8486638391257812106487756118005899929206559764024760867909861065837210171584253486428255535562172797
8087524927419514988133189010270748540131244546425970929428585158043675385979294934983548991320978707
9643410175774190118837945476362354068164746267420381161720513440838355758850658162106816819321526271
65...

>> No.1201872 [View]
File: 119 KB, 903x986, Riemann..jpg [View same] [iqdb] [saucenao] [google]
1201872

>Q: Why can't parallel lines intersect?
>A: Because we are using a Euclidean metric

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