I''m impressed, man!
ToohrVyk
Other math puzzles
How about this one:
What are the real value(s) of all the complex numbers n that satisfy the following:
0 = The sum from k = 1 to infinity of 1/(k^n). (Like a p-series where p = n and n is a complex number.)
What are the real value(s) of all the complex numbers n that satisfy the following:
0 = The sum from k = 1 to infinity of 1/(k^n). (Like a p-series where p = n and n is a complex number.)
LOL :D.
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!
It is 1/2. Now that I answered, why don''t you post your solution to it
!
20: ...Qg2 ++
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!
It is 1/2. Now that I answered, why don''t you post your solution to it
! 20: ...Qg2 ++
Author
Given a vectorial space, a scalar product for this space, and two vectors u != v from this space, prove that if you take ANY two vectors m, m'' for which ||u-m|| = ||v-m|| and ||u-m''|| = ||v-m''||, then ( u-v | m - m'' ) = 0. (IE u-v and m-m'' are orthogonal).
ToohrVyk
ToohrVyk
quote:Original post by GaulerTheGoat
LOL :D.
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!
It is 1/2. Now that I answered, why don''t you post your solution to it
20: ...Qg2 ++
Well techinally that hasn''t been proven yet (has it?)
.
quote:Original post by The Hereticquote:Original post by GaulerTheGoat
LOL :D.
Who are you kidding, AP? I will go wake up Riemann. He will figure this out! Hahahahahahahahaha!
It is 1/2. Now that I answered, why don''t you post your solution to it
20: ...Qg2 ++
Well techinally that hasn''t been proven yet (has it?)
I have seen in one of my old magazines that a theorom proving program proved it with a certain probability. I didn''t understand the article but I will either find a link to it or try to copy it when I get home. It had something to do with Hyper-geometric functions which I also don''t get. I will post later, then you can decide
. 20: ...Qg2 ++
quote:Original post by ToohrVyk
Given a vectorial space, a scalar product for this space, and two vectors u != v from this space, prove that if you take ANY two vectors m, m'' for which ||u-m|| = ||v-m|| and ||u-m''|| = ||v-m''||, then ( u-v | m - m'' ) = 0. (IE u-v and m-m'' are orthogonal).
ToohrVyk
Are the scalars real numbers, complex, or just a generic field? Is scalar product the same as a Hermitian inner product?
I just started getting more interested in vector spaces. Tell me how I am doing :
Vector differences are independant of the origin, so choose m . So
||u -m || = ||v -m || implies that
||u || = ||v ||. Multiplying out the norms
(u -m )•(u -m ) =
(v -m )•(v -m ) or
u •u -2u •m +m •m =
v •v -2v •m +m •m or
u •m = v •m .
The other equation gives similarly,
u •m'' = v •m'' .
Subtracting these two and factoring out the u on the LHS and the v on the RHS, and then subtracting the RHS and factoring out the m -m'' gives
(u -v )•(m -m'' ) = 0.
20: ...Qg2 ++
:Vector differences are independant of the origin, so choose m . So
||u -m || = ||v -m || implies that
||u || = ||v ||. Multiplying out the norms
(u -m )•(u -m ) =
(v -m )•(v -m ) or
u •u -2u •m +m •m =
v •v -2v •m +m •m or
u •m = v •m .
The other equation gives similarly,
u •m'' = v •m'' .
Subtracting these two and factoring out the u on the LHS and the v on the RHS, and then subtracting the RHS and factoring out the m -m'' gives
(u -v )•(m -m'' ) = 0.
20: ...Qg2 ++
quote:Original post by GaulerTheGoat
I just started getting more interested in vector spaces. Tell me how I am doing
Vector differences are independant of the origin, so choose m . So
||u -m || = ||v -m || implies that
||u || = ||v ||. Multiplying out the norms
(u -m )•(u -m ) =
(v -m )•(v -m ) or
u •u -2u •m +m •m =
v •v -2v •m +m •m or
u •m = v •m .
The other equation gives similarly,
u •m'' = v •m'' .
Subtracting these two and factoring out the u on the LHS and the v on the RHS, and then subtracting the RHS and factoring out the m -m'' gives
(u -v )•(m -m'' ) = 0.
20: ...Qg2 ++
I''m not sure where you''re getting
||u || = ||v ||
but you don''t need it anyway. If you subtract the two equations at the end those two terms drop out because they are in both equations:
This:
u •u -2u •m +m •m =
v •v -2v •m +m •m
And this:
u •u -2u •m'' +m'' •m'' =
v •v -2v •m'' +m'' •m''
This topic is closed to new replies.
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