(Not sure what number we are really on, but I'll say 56.)
Let's count to infinity!
Re: Let's count to infinity!
by SpikedMath » Tue Feb 22, 2011 10:12 am
(Not sure what number we are really on, but I'll say 56.)
Math - It's in you to give.
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Re: Let's count to infinity!
by Ardilla » Tue Feb 22, 2011 11:57 am
Spoiler! :
Clearly every even integer greater than 2 can be expressed as the sum of two primes.
I have discovered a truly wonderful proof of this proposition, but the signature is too small to contain it.
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Re: Let's count to infinity!
by DeathRowKitty » Tue Feb 22, 2011 12:00 pm
Pretty sure that's what we were supposed to be up to.
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Re: Let's count to infinity!
by WhoDatMath » Thu Feb 24, 2011 12:08 pm
Last edited by WhoDatMath on Mon Feb 28, 2011 9:31 am, edited 1 time in total.
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Re: Let's count to infinity!
by DeathRowKitty » Thu Feb 24, 2011 12:57 pm
(That should be the 17th prime, not the 16th)
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Re: Let's count to infinity!
by bmonk » Thu Mar 03, 2011 6:16 pm
[(4^4)/4 - .4]
(That outer set of square brackets is the greatest integer function)
So, nobody wants to get to 64?
63 is pretty far from infinity...
(That outer set of square brackets is the greatest integer function)
So, nobody wants to get to 64?
63 is pretty far from infinity...
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Re: Let's count to infinity!
by bmonk » Sat Apr 23, 2011 5:49 pm
Alas, nobody is interested in the (5! / √4) + 3 + 2 - 1-th number???
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Re: Let's count to infinity!
by Ekuurh » Tue May 03, 2011 11:40 am
the only number N that can be represented as N=(a^b)^(b^a)=(b^a)^(a^b)
There was an open question which i have asked many proffesors and fellow mathematicians, but none were able to answer:
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
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Re: Let's count to infinity!
by SpikedMath » Tue May 03, 2011 3:51 pm
In hexadecimal, this number is 42.
Math - It's in you to give.
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Re: Let's count to infinity!
by bmonk » Thu May 05, 2011 5:52 pm
The US route from Chicago to LA, plus one.
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Re: Let's count to infinity!
by Ekuurh » Wed May 18, 2011 5:56 am
let f(n) be the n-th Fibonacci number (with f(0)=1, f(1)=1, f(n+2) = f(n+1)+f(n)):
The only number that can be represented as f(n)*f(f(n)^f(n+1))=f(1+f(n)^f(n+1))+f(f(n)*f(n+1))
I'll make it easier for you:
n=2.
The only number that can be represented as f(n)*f(f(n)^f(n+1))=f(1+f(n)^f(n+1))+f(f(n)*f(n+1))
I'll make it easier for you:
n=2.
There was an open question which i have asked many proffesors and fellow mathematicians, but none were able to answer:
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
- Ekuurh
- Elementary School
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- Joined: Tue Apr 19, 2011 12:28 pm
Re: Let's count to infinity!
by Ekuurh » Mon May 23, 2011 7:42 am
[img]http://spikedmath.com/cgi-bin/mimetex.cgi?{8}%5Cchoose{4}[/img]
There was an open question which i have asked many proffesors and fellow mathematicians, but none were able to answer:
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
- Ekuurh
- Elementary School
- Posts: 14
- Joined: Tue Apr 19, 2011 12:28 pm
Re: Let's count to infinity!
by Ekuurh » Mon May 23, 2011 7:42 am
Ekuurh wrote:[img]http://spikedmath.com/cgi-bin/mimetex.cgi?{8}%5Cchoose{4}[/img]
damn, it ain't working, just go to
http://spikedmath.com/cgi-bin/mimetex.cgi?{8}%5Cchoose{4}
There was an open question which i have asked many proffesors and fellow mathematicians, but none were able to answer:
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
Is there a function f:R->R (not necessarily continuous on ALL of R) which:
for every linear function g(x)=m*x+n, there is a number X for which g is the tangent of f in? (for each x g(x)=f(X)+f'(X)(x-X))
i think not.
(i almost solved it - using a small set theory tweak and a big topology tweak)
- Ekuurh
- Elementary School
- Posts: 14
- Joined: Tue Apr 19, 2011 12:28 pm
Re: Let's count to infinity!
by bmonk » Mon May 30, 2011 10:00 pm
I'm slipping!
How about: [(100* √2)+1] / 2
You were right--forgot the divide by 2. Alas--I'm one of those mathematicians who can't calculate, and even find counting a chore.
How about: [(100* √2)+1] / 2
You were right--forgot the divide by 2. Alas--I'm one of those mathematicians who can't calculate, and even find counting a chore.
Last edited by bmonk on Wed Jun 01, 2011 10:19 am, edited 1 time in total.
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Re: Let's count to infinity!
by SpikedMath » Wed Jun 01, 2011 8:36 am
I think you need a division by 2 bmonk.
I'll do an easy one:
I'll do an easy one:
Math - It's in you to give.
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Re: Let's count to infinity!
by Ardilla » Mon Jun 06, 2011 8:42 am
74
Spoiler! :
Clearly every even integer greater than 2 can be expressed as the sum of two primes.
I have discovered a truly wonderful proof of this proposition, but the signature is too small to contain it.
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Ardilla - High School
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