Mathematics Thread

[Ickathu]

Who wants to play 24?
http://www.speedsolving.com/forum/showthread.php?40791-Twenty-Four-Marathon

 

[brandbest1]

Someone explain to me the purpose of the dot product and cross product?

 

[vcuber13]

a cross b gives the vector perpendicular to a and b (right hand rule), with magnitude equal to the area of the parallelogram formed by a and b
a dot b is equal to magnitude of a times the magnitude of b times cos theta

 

[brandbest1]

a cross b gives the vector perpendicular to a and b (right hand rule), with magnitude equal to the area of the parallelogram formed by a and b
a dot b is equal to magnitude of a times the magnitude of b times cos theta

No, I meant what the purpose of these are. I saw that it it used somewhere in the law of cosines or something like that.

And someone solve this question:
Find the ordered triple (x,y,z) of positive integers with x<y<z that satisfies xyz+xy+yz+xz+x+y+z=384.

 

[vcuber13]

if vector c is vector a minus vector b then: (where vectors a, b, and c are the sides of a triangle)
\(
c \cdot c = (a-b) \cdot (a-b)
\)
\(
= a \cdot a - a \cdot b - b \cdot a + b \cdot b
\)
\(
=a^2 - 2 a \cdot b + b^2
\)
\(
c^2 = a^2 + b^2 - 2ab Cos \theta
\)

 

[ben1996123]

And someone solve this question:
Find the ordered triple (x,y,z) of positive integers with x<y<z that satisfies xyz+xy+yz+xz+x+y+z=384.

4,6,10

also tudae i got mein results for c3 and fp1. 93/100, 100/100. in c3 there was a stupid pruf question so i prollaby got that "wrong". pruf 3^x never ends in 5 for possitif integer x. apparently noticing the pattern in the last digit: 3,9,7,1,3,9,7,1,blablabla is a rigorous enuf pruf but prufing the fundamental theorem of arithmetic and saying that if 3^x never ends in 5 then 3^x (mod 5) is nevar 0, wich it isnt because 3^x = 3*3*3*3*blablabla and 3 is prime and 3 is not 5 is not a rigorous enuf pruf.

 

[ben1996123]

ok so, i have disproven everything

[15:16:54] Dashie: ok so
[15:17:00] Dashie: sgn(0) is 0 rye ?
[15:17:19] 5BLD: yes
[15:17:28] Dashie: and sgn(x) = e^iarg(x)
[15:17:50] Dashie: so sgn(0) = e^undefined
[15:17:53] Dashie: LOL !
[15:18:06] Dashie: e^iundefined*
[15:18:22] Dashie: there four iundefined = ln(0) so undefined = -iln(0)
[15:18:36] Dashie: = ∞i
[15:18:39] Dashie: wol
[15:18:45] 5BLD: nise reasoning
[15:18:51] 5BLD: cuz inf=inf in all cases rye
[15:18:55] Dashie: shore LOL !
[15:19:01] Dashie: 1/0 = ∞i
[15:19:07] Dashie: diwideing by inf
[15:19:11] Dashie: 1/(0∞) = i
[15:19:16] Dashie: 1/undefined = i
[15:19:20] Dashie: undefined = -i
[15:19:36] Dashie: undefined = -i = -iln(0) = ∞i
[15:19:42] Dashie: therefore ln(0) = 1 and ∞ = -1
[15:19:47] Dashie: therefore e = 0
[15:19:48] Dashie: LOL !

 

[5BLD]

First of all, prove theres always solutions to kx²+1=y² where k,x,y ∈ N for all k

then describe the growth of the smallest x solution to
(2ⁿ-1)x²+1=y²
where n,x,y ∈ N
as n increases

 

[ben1996123]

First of all, prove theres always solutions to kx²+1=y² where k,x,y ∈ N for all k

then describe the growth of the smallest x solution to
(2ⁿ-1)x²+1=y²
where n,x,y ∈ N
as n increases

noob y dnt u just make a progrm 2 chek all da numbers

oarso, hear is mein "pruf" of fermats last theorem

 

[redbeat0222]

Why you no join maff team?

 

[5BLD]

noob y dnt u just make a progrm 2 chek all da numbers

yeah, cuz we can check every natural number

 

[ben1996123]

Why you no join maff team?

me? csch i got no maff team to joixn

yeah, cuz we can check every natural number

yes u can computer's r instante rye

 

[5BLD]

Calculator/Math Thread

just sorve my prahblem moron

Tis like saying cuz its tru for n<100 tis tru for all n.
i mean like look at this 'pruf' (related to my problem btw).

Conjecture: there are no numbers n where 991n^2+1 is a square number.

Pruf: You can test to like, quadrillions (iirc) and you wont find one

Atchiry theres quite a large answer as the smallest x which works.

 

[vcuber13]

nice variable consistency

 

[Ickathu]

Proving that the product of the positive divisors of a perfect cube is a perfect cube. How's this proof?

Spoiler

 

[vcuber13]

or you could prove it like this

if you have a^3 its divisors are a and a^2, if a is factored and multiplied the factors will always have a degree of 3 and thus is a cube

 

[kunparekh18]

Re: Calculator/Math Thread

or you could prove it like this

if you have a^3 its divisors are a and a^2, if a is factored and multiplied the factors will always have a degree of 3 and thus is a cube

Yeah, this is way simpler

Sent from my A75 using Tapatalk 2

 

[redbeat0222]

2+2 equals 4. <_<
>_>
Q.Q

 

[ben1996123]

2+2 equals 4. <_<
>_>
Q.Q

pruf

 

[ben1996123]

oo this is koo, everyjuan knows that \( e^{i\frac{\tau}{2}}=-1 \), but apparencely \( (\frac{\tau}{2})^{ie}\approx-1 \) too