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/sci/ - Science & Math

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>> No.8581780 [View]

>>8580634
your inferiority is showing

>> No.8581718 [View]

>>8580717
positive integers A, B, C are all
smaller than your value for C

>> No.8580633 [View]

>>8578851
Correct, I transposed the 1 and 7 because fatigue.

>> No.8580631 [View]

>>8578416
>>8578538
>homework
No, when I wrote that I found this written in the back of an old book (Mathematical Handbook by Murray R.Spiegel) that was true, unlike what you wrote.

>> No.8578554 [View]

>>8578535
oops

>> No.8578376 [View]

>>8578202
>a white couple with no black ancestors
There is no such couple.

>> No.8578371 [View]

>>8578000
>What is Mr Hoffengoblins endgame here?
He is not a chessplayer, he has no endgame.

>> No.8578365 [View]

>>8577399
>>8577600
get over your IQ fixation fgt pls
protip: defend your weakness, not your strength

>> No.8578359 [View]
File: 140 KB, 640x480, pi-640-480.jpg [View same] [iqdb] [saucenao] [google]
8578359

I found this written in the back of an old Math book today:
[math] \displaystyle \pi \approx \frac{208341}{66371}= \frac{A}{17}+ \frac{B}{47}+ \frac{C}{83}[/math]
and it appears to be within one-half part per trillion of the actual value. Find the values of A, B, and C, if you're hard enough.

>> No.8573865 [View]
File: 22 KB, 733x363, probzerosuccess.png [View same] [iqdb] [saucenao] [google]
8573865

>>8573817
>my friend says 13%
he's right

>> No.8573853 [View]

>>8570647
took me a while to figure them out,
looks a bit like the graph of
[math]xe^{-x}[/math]
thanks

>> No.8570651 [View]

>>8570643
>I said basically infinity
therefore basically retarded
gtfo fgt pls

>> No.8570647 [View]

>>8570630
>>8570631
...what part of x ≥ 0 did we miss, Anon?
...and those graphs are totally whack.

>> No.8570619 [View]

>>8570606
>divided by infinity
division is a binary operation on numbers
infinity is not a number
therefore no division by infinity

>> No.8570616 [View]
File: 2 KB, 540x260, longdivision.png [View same] [iqdb] [saucenao] [google]
8570616

if 0.999...≠ 1 then 1 – 0.999... > 0
so call it ε = 1 – 0.999...
and divide 1 by ε via long-division;
you get a series that diverges,
therefore ε = 0 , QED

>> No.8570596 [View]

missed it by *that* much
[math]n \epsilon \mathbb{N}[/math]

>> No.8570587 [View]
File: 10 KB, 250x250, firework.png [View same] [iqdb] [saucenao] [google]
8570587

I know this function F(x) converges,
because terms of [math]e^x[/math] converge,
and [math]L_n(x)[/math] is decreasing, duh
but how do I graph this doggie?
[math]\displaystyle F(x)= \lim_{n \rightarrow \infty}F_n(x)=e^{-x}[L_1(x)+xL_2(x)+\frac{x^2}{2!}L_3(x)+ \cdots+\frac{x^n}{n!}L_{n+1}(x)][/math]
where [math]n \epsilon \mathbb{N}[/math]
and [math]x \geq 0[/math]
and [math]L_0(x)=x[/math]
and [math]L_n(x)= \ln(L_{n-1}(x)+1)[/math]

>> No.8545091 [View]

>>8545076
>division is multiplication
operation-A yielding the same result
as operation-B does not imply A is B

>> No.8502927 [View]

>>8502793
>I looked at various algebraic and analytical proofs
no, you did not

>> No.8310726 [View]

>>8310723
surely there's an easier, quicker way

>> No.8310724 [View]

>>8310696
>no harmful mining and factory work
>required for nuclear energy
fgt pls

>> No.8310689 [View]
File: 832 B, 249x51, belphegor-prime.gif [View same] [iqdb] [saucenao] [google]
8310689

How would one go about proving the number
[math]10^{30}+666 \times 10^{14}+1[/math]
is prime?

>> No.8308167 [View]

>>8308156
>I had no idea
yes

>> No.8308163 [View]

>>8307182
wat

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