vcuber13
Veteran
dont like pis eh?
Mathematics Thread
- Thread starter IanTheCuber
- Start date
CarlBrannen
Veteran
Looking for an excuse to try out the math mode, here's a little puzzle: reduce: \( i^i \)
5BLD
Legend
ben1996123
Banned
some of these seem pretty interesting
AB = 2x, BC = 4-x²
x = AB/2
BC = 4-(AB)²/4
ABCD is a square when 2x = 4-x² and x>0
x² + 2x - 4 = 0
x = √5-1
side length is 2x = 2√5-2
need to find the red area
x = 1/2 because its the radius of the circle and the diameter is 1
y² + y² = x²
2y² = 1/4
\( y=\frac{1}{2\sqrt{2}} \)
\( 2y=\frac{1}{\sqrt{2}} \)
area of the smaller square = (2y)² = 1/2
red area = area of circle - area of square = \( \pi x^2-1/4=\frac{\pi-1}{4} \)
dunfeeerye doing anymore now
AB = 2x, BC = 4-x²
x = AB/2
BC = 4-(AB)²/4
ABCD is a square when 2x = 4-x² and x>0
x² + 2x - 4 = 0
x = √5-1
side length is 2x = 2√5-2
need to find the red area
x = 1/2 because its the radius of the circle and the diameter is 1
y² + y² = x²
2y² = 1/4
\( y=\frac{1}{2\sqrt{2}} \)
\( 2y=\frac{1}{\sqrt{2}} \)
area of the smaller square = (2y)² = 1/2
red area = area of circle - area of square = \( \pi x^2-1/4=\frac{\pi-1}{4} \)
\( a(+)b=\frac{1}{\frac{1}{a}+\frac{1}{b}}=\frac{ab}{a+b} \)
\( a(-)b=\frac{1}{\frac{1}{a}-\frac{1}{b}}=\frac{ab}{b-a} \)
\( a(-)b=\frac{1}{\frac{1}{a}-\frac{1}{b}}=\frac{ab}{b-a} \)
\( a(-)b = \frac{b}{a} \)
\( \frac{ab}{b-a}=\frac{b}{a} \)
\( a^2b=b^2-ab \)
\( a^2b+ab-b^2=0 \)
\( a^2+a-b=0 \)
\( b=a^2+a \)
\( \frac{ab}{b-a}=\frac{b}{a} \)
\( a^2b=b^2-ab \)
\( a^2b+ab-b^2=0 \)
\( a^2+a-b=0 \)
\( b=a^2+a \)
\( a(+)b=2a(-)b \)
\( \frac{ab}{a+b}=\frac{2ab}{b-2a} \)
\( ab(b-2a)=2ab(a+b) \)
\( ab^2-2a^2b=2a^2b+2ab^2 \)
\( b-2a=2a+2b \)
\( b=-4a \)
\( \frac{ab}{a+b}=\frac{2ab}{b-2a} \)
\( ab(b-2a)=2ab(a+b) \)
\( ab^2-2a^2b=2a^2b+2ab^2 \)
\( b-2a=2a+2b \)
\( b=-4a \)
\( a(-)b=b \)
\( \frac{ab}{b-a}=b \)
\( ab=b^2-ab \)
\( 2ab=b^2 \)
\( b=2a \)
\( \frac{ab}{b-a}=b \)
\( ab=b^2-ab \)
\( 2ab=b^2 \)
\( b=2a \)
Only 1 solution exists when y = x + c is a tangent to \( y=2^x \) and \( \frac{d}{dx}2^x=1 \)
\( 2^xln(2)=1 \)
\( 2^x=\frac{1}{ln(2)} \)
\( xln(2)=ln(\frac{1}{ln(2)})=-ln(ln(2)) \)
\( x=\frac{-ln(ln(2))}{ln(2)} \)
The tangent has gradient 1 and goes through \( (\frac{-ln(ln(2))}{ln(2)},\frac{1}{ln(2)}) \)
\( y-\frac{1}{ln(2)}=x-\frac{-ln(ln(2))}{ln(2)} \)
\( y=x+\frac{1+ln(ln(2))}{ln(2)} \)
\( c=\frac{1+ln(ln(2))}{ln(2)} \)
\( 2^xln(2)=1 \)
\( 2^x=\frac{1}{ln(2)} \)
\( xln(2)=ln(\frac{1}{ln(2)})=-ln(ln(2)) \)
\( x=\frac{-ln(ln(2))}{ln(2)} \)
The tangent has gradient 1 and goes through \( (\frac{-ln(ln(2))}{ln(2)},\frac{1}{ln(2)}) \)
\( y-\frac{1}{ln(2)}=x-\frac{-ln(ln(2))}{ln(2)} \)
\( y=x+\frac{1+ln(ln(2))}{ln(2)} \)
\( c=\frac{1+ln(ln(2))}{ln(2)} \)
dunfeeerye doing anymore now
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ben1996123
Banned
okso what is i^j and i^^i?
ithink i^j = -k but it seems wrong because it seems like it should be transcal/transcendal/transcendental/transcendentenal/transcendentenendal
ithink i^j = -k but it seems wrong because it seems like it should be transcal/transcendal/transcendental/transcendentenal/transcendentenendal
Ickathu
Veteran
what is j and k and stuff? I've heard of them, but I can't find what they are exactly...
Noahaha
blindmod
ben1996123
Banned
stuf
z=a+bi+cj+dk where a,b,c,d are real numbers
i^2 = j^2 = k^2 = -1
i*j = k
i*k = -j
j*i = -k
j*k = i
k*i = j
k*j = -i
in the set of quaternions, there are iniffity sqruts of -1 which is cool
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Ickathu
Veteran
cool stuff...
Is it possible for a number in one base to be written the same way in another base? I'm thinking that it's not possible for them to ever be the same (except with single digit numbers, ofc).
Here's my "proof" for 2 digit numbers:
Call the 2 digit number in base m BA with B=/=0
BA_m can be rewritten in base 10 as:
B*m+A
Assume that BA_m = BA_n
BA_n can be rewritten in base 10 as B*n+A
BA_m = BA_n --> B*m+A=B*n+A
using algebra:
B*m=B*n
m=n
contradiction, so BA_m =/= BA_n, if m=/=n and B =/= 0
I started working on a similar proof for 3 digit numbers, but I got stuck when I got to this (A is the base^0 digit, B is base^1 digit, C is base^2, etc...):
B+C(n+m)+D(n^2+nm+m^2)=0
I could explain how I got to that, but I don't think it's really going to work.
Is it possible for a number in one base to be written the same way in another base? I'm thinking that it's not possible for them to ever be the same (except with single digit numbers, ofc).
Here's my "proof" for 2 digit numbers:
Call the 2 digit number in base m BA with B=/=0
BA_m can be rewritten in base 10 as:
B*m+A
Assume that BA_m = BA_n
BA_n can be rewritten in base 10 as B*n+A
BA_m = BA_n --> B*m+A=B*n+A
using algebra:
B*m=B*n
m=n
contradiction, so BA_m =/= BA_n, if m=/=n and B =/= 0
I started working on a similar proof for 3 digit numbers, but I got stuck when I got to this (A is the base^0 digit, B is base^1 digit, C is base^2, etc...):
B+C(n+m)+D(n^2+nm+m^2)=0
I could explain how I got to that, but I don't think it's really going to work.
CarlBrannen
Veteran
Zero is always written as "0" no matter what base, and one is always written "1". Other than that, no solutions.
5BLD
Legend
Calculator/Math Thread
He said except for single digit numbers AND wanted a pruf
He said except for single digit numbers AND wanted a pruf
ben1996123
Banned
okso its tru because the only real solution to a^x = b^x where x is real and a!=b is x=0 => a^0=b^0 => 1=1 => sniggledigit
Ickathu
Veteran
Cool, also I asked some other people about this and they said to use that if m>n, then m^k>n^k for k>0, so therefore a*m^k =/= a*n^k
TheNextFeliks
Veteran
sgn(0)=:fp
ben1996123
Banned
sgn(0) = 0 moron
also am pretty sure i^j = -k now. i just need to prove that ln(a^b)=bln(a) for some quaternionstuf
okso ln(x) = (x-1) - (x-1)^2/2 + (x-1)^3/3 - dotdotdot when -2<x<0 fugginel duncare ill assume its trú for quaternions too if tis tru for x = 1+i
ln(i+1) = ln(2)/2 + ipi/4
ln(i+1) = i - i^2/2 + i^3/3 - i^4/4 + i^5/5 - i^6/6 + i^7/7 - i^8/8 + ... = i + 1/2 - i/3 - 1/4 + i/5 + 1/6 - i/7 - 1/8 + ...
= (1/2 - 1/4 + 1/6 - 1/8 + ...) + i(1 - 1/3 + 1/5 - 1/7 + ...)
= 1/2(1 - 1/2 + 1/3 - 1/4 + ...) + i(1 - 1/3 + 1/5 - 1/7 + ...)
= 1/2 ln(2) + ipi/4
= tru, therefore ln(x+1) qonverges for all quaternions if |q|<1 or |q|=1 because I gave 1 example rye
ln(x+1)-ln(x) = ln(1/x + 1) rye so ln(1+i^j)-ln(1+i^-j) = ln(i^j)
ill assume moar stuff that maeby isnt true but whoqaers because maff is about making stuff up rye
ln(1+i^j) = i^j - i^2j/2 + i^3j/3 - i^4j/4 + i^5j/5 - ... = i^j + 1^j /2 - i^j/3 - 1^j /4 + i^j/5 - ... = i^j(1-1/3+1/5-...) + 1^j(1/2-1/4+1/6-...)
lettuce assume that 1^j = 1
ln(1+i^j) = i^j pi/4 + ln(2)/2
okso lettuce assume that ln(1+q) = qpi/4 + ln(2)/2 because I gave 2 examples rye
ln(1+i^-j) = i^-j pi/4 + ln(2)/2
so ln(i^j) = ln(1+i^j)-ln(1+i^-j) = (i^j pi/4 + ln(2)/2) - (i^-j pi/4 + ln(2)/2) = pi/4 (i^j - i^-j)
i^j - i^-j = i^-j(i^2j-1) = i^-j(-1^j - 1)
-1^j = e^ipij = e^kpi = -1
i^-j(-2) = -2i^-j = 2i^j maeby
ln(i^j)= pi/4 (2i^j) = i^j pi/2
ln(i) = i pi/2
jln(i) = jipi/2 fugginel thats not i^j pi/2 it was supposed to work perfectly stupid maff
okso i^j = e^(i^j pi/2) = fugginel its still got i^j inside of it
fugginel why dose stuff always work for euler but not for me
mark49152
Premium Member
That made me LOL
5BLD
Legend
Right guys so ive been trying to work out the principal value of (a^b)^c
but first. what's the principal value of ln(a^b)?
We define the principal natural logs to have an imaginary part between -pi and pi but only by adding multiples of 2pi
Re(ln(a^b))+Im(ln(a^b)) mod 2pi gives you an imaginary part between -2pi and 2pi and we need to somehow add either -pi or pi to get it right
So this is what I'm left with:
b*ln(a)+2ipi(floor(b*ln(a)/(2pi))) (+/-pi???)
hmmmh
Edit: I'm being siwwy, i cant just add or take away pi randomly
Needs to "line up" with taking away multiples of 2pi
but first. what's the principal value of ln(a^b)?
We define the principal natural logs to have an imaginary part between -pi and pi but only by adding multiples of 2pi
Re(ln(a^b))+Im(ln(a^b)) mod 2pi gives you an imaginary part between -2pi and 2pi and we need to somehow add either -pi or pi to get it right
So this is what I'm left with:
b*ln(a)+2ipi(floor(b*ln(a)/(2pi))) (+/-pi???)
hmmmh
Edit: I'm being siwwy, i cant just add or take away pi randomly
Needs to "line up" with taking away multiples of 2pi
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kunparekh18
Veteran
Do they really teach you all of that in school? :O What grade/standard are you in?
5BLD
Legend
Calculator/Math Thread
Lol youfink I'd talk about school maths here
Im 15 but i know a bit higher maths just for fun. I haven't worked it out yet, its silly.
3pi/2 rad =>-pi/2
-3pi/2 rad =>pi/2
Hmmh gah its so obvious but cant work it out
Log(a^b)=> b log a - 2ipi floor (1/2 - Im(b log a)/2pi) seems to do the trick but hmmmh
Lol youfink I'd talk about school maths here
Im 15 but i know a bit higher maths just for fun. I haven't worked it out yet, its silly.
3pi/2 rad =>-pi/2
-3pi/2 rad =>pi/2
Hmmh gah its so obvious but cant work it out
Log(a^b)=> b log a - 2ipi floor (1/2 - Im(b log a)/2pi) seems to do the trick but hmmmh
kunparekh18
Veteran
I'm 15 too, but yeah, I can't understand some of the higher maths. Why don't we have a question-answer sort of thing. My first question:
If the sum of m terms of an arithmetic progression is the same as the sum of it's n terms show that the (m+n)th term (of the AP) is the negative of the first term.
If the sum of m terms of an arithmetic progression is the same as the sum of it's n terms show that the (m+n)th term (of the AP) is the negative of the first term.
