Well done man, I'll say that was blagged! Once you do the even case, you will realize that it is essentially the same as the even case!(Original post by henpen)
Solution 286
For even n,
For odd n, I'll try to generalise 24 later, but I need to sleep now.
Hence, we have: which implies that the answer was
which I suppose can be seen as yet another reason to introduce as being equal to .
Does anyone have any strong views on this? Is more arbitrary than ?
This is essentially a consequence of the multiplication theorem in the case where we let .

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 28072013 17:03

Smaug123
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 28072013 17:28

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 28072013 17:49
(Original post by Smaug123)
The only place I've seen where pi was more natural than tau was in the integral from inf to inf of exp(x^2). Then we have sqrt(pi).
Problem 290*
Evaluate in the case where are, in 3dimensional Euclidean space, the angles respectively between the line connecting the origin with any arbitrary point and any arbitrarily chosen orthonormal basis.
Show that this can be seen as a generalization of Pythagoras' Theorem (or, equivalently, a means of alternative proof).
Can the result be generalized to ndimensional Euclidean space where the coordinates lie in ? (bonusonly ***)Last edited by Jkn; 28072013 at 18:36. Reason: typo 
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 28072013 18:30
(Original post by Jkn)
'The only place'? There are hundreds! The most convincing is probably that it's the angle in a straight line (which, to me, seems less arbitrary than the angle in a circle).
(quibble with the wording of your Problem 290: "any three orthonormal bases" should be "any three orthonormal vectors" or "any orthonormal basis") 
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 2005
 28072013 18:38
(Original post by Smaug123)
(quibble with the wording of your Problem 290: "any three orthonormal bases" should be "any three orthonormal vectors" or "any orthonormal basis")
Btw, why have you not been solving my problems? 
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 2006
 28072013 18:46
(Original post by Jkn)
Cheers. How is "orthonormal bases" not correct wording? I essentially just mean axes as the application of vectors to the problem is not a necessity.
Btw, why have you not been solving my problems?
Bleugh, I've got homework I'll have a go at some 
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 2007
 28072013 18:53
(Original post by Jkn)
What I've been doing in some cases is trying to think of some myself; often generalisations of other problems I have tried/found Such as...
Problem 96*
Let x, y and a be positive integers such that . Given that where b is a positive integer, find, in terms of a and b, the number of possible pairs that satisfy the equation.
Not taken much brainpower over that, though, so I might be wrong :P 
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 2008
 28072013 20:15
(Original post by Jkn)
Well done man, I'll say that was blagged!
Wait, I may have gotten that wrong earlier in my sleepinduced stupor, it may very well generalise.Last edited by henpen; 28072013 at 20:34. 
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 28072013 20:29
(Original post by Jkn)
Problem 290*
Evaluate in the case where are, in 3dimensional Euclidean space, the angles respectively between the line connecting the origin with any arbitrary point and any arbitrarily chosen orthonormal basis.
Show that this can be seen as a generalization of Pythagoras' Theorem (or, equivalently, a means of alternative proof).
Can the result be generalized to ndimensional Euclidean space where the coordinates lie in ? (bonusonly ***)
alp
Take a vector such that . is its projection onto the axis, onto the axis ... .
Thus is its projection onto the plane, and is its projection onto the '3D space' (what word do I use here, ?), and is its projection onto the 'nD space'. But this projection is a line of length , so and .
I hope you understood what I meant. Could you help formalise the bits where I was hopelessly off with regards to the mathematical language?Last edited by henpen; 28072013 at 20:33. 
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 28072013 20:33
(Original post by Smaug123)
An orthonormal basis is a set of orthogonal unit vectors  so "any three arbitrarily chosen ON bases" means "any three sets of three orthogonal unit vectors". I understood what you meant, though
Bleugh, I've got homework I'll have a go at someLast edited by henpen; 28072013 at 20:37. 
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 28072013 20:35
(Original post by Smaug123)
Is this not just ? Since the expression is , so a+x is square; there are sqrt(a+b) square numbers available for the a+x term.
Not taken much brainpower over that, though, so I might be wrong :P
Why not try some of my fiendish integral/EM/Gamma function problems? (I'm hoping people will solve them so that I wont feel bad posting more ). Why not post some yourself?
I'm getting quite in to my vectors now so look forward to some questions of that type
(Original post by henpen)
Sort of, I can't seem to generalise result 24 when the '1006' is odd: the argument of isn't equal to 0, so the result isn't as simple.
Wait, I may have gotten that wrong earlier in my sleepinduced stupor, it may very well generalise.
Actually, what I found most trilling is that we now have a rather strange alternate solution to problem 24 using the everpowerful gamma function (i.e. multiplication theorem). Waaaay too hungover to type it up though.. : 
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 2012
 28072013 20:50
(Original post by henpen)
'Solution' 290
Also, your last part is right but it seems as though, at this point, it is merely a conjecture based on the initial geometric argument. Perhaps it would be better to use a more abstract algebraic argument based on a more sophisticated (and more abstract) understanding of vector principles?
I hope you understood what I meant. Could you help formalise the bits where I was hopelessly off with regards to the mathematical language? 
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 2013
 28072013 21:05
(Original post by Jkn)
For the sake of clear explanation, it would probably be helpful to note why it is that the choice of such a vector does not lose you any generality. Alternatively, you could just consider a vector of arbitrary length?
Also, your last part is right but it seems as though, at this point, it is merely a conjecture based on the initial geometric argument. Perhaps it would be better to use a more abstract algebraic argument based on a more sophisticated (and more abstract) understanding of vector principles?
Smaug seemed far more wellversed with vector terminology etc.. so I will leave it to him to formalize your language (seems fine to me ).
Expand the vector componentwise,:
and its modulus:
But note that by the vector's expansion, and by the definition of , and thus , giving

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 28072013 21:46
(Original post by henpen)
Can you not also have as orthonormal basis vectors? I hear that definition's got something to with them being orthogonal under integration (between certain limits).
They are orthogonal under the innerproduct defined by
edit: Problem 291*
Prove the above statement that sines and cosines are orthogonal under the inner product given. This is to show the integral of , , or , are zero if and only ifLast edited by FireGarden; 28072013 at 21:52. 
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 28072013 21:49
(Original post by henpen)
'Solution' 290 
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 28072013 21:51
I eliminated the need for geometry by interpreting cosine by its place in the dot product (I prefer to 'get on with' the algebra in questions of this sort).
Post some questions bro? 
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 2017
 28072013 21:54
(Original post by FireGarden)
You can; they are the basis vectors of functions used when dealing with fourier series.
They are orthogonal under the innerproduct defined by 
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 28072013 21:58
(Original post by Jkn)
I remember that from QM! (one function should be conjugated btw, unless their codomain is real) 
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 2019
 28072013 22:06
Problem 292***
Group summary:
Spoiler:ShowI define a "group" to be a set X, and an operation +, such that the following four axioms are true:
1) a+b is in X for every a,b in X
2) there is an element, e (the identity), such that a+e = a = e+a for all a in X
3) every element a in X has a corresponding element z in X with a+z = z+a = e
4) (a+b)+c = a+(b+c) for all a,b,c in X.
So the set "the integers, with +" is a group:
1) a+b is an integer for all integers a,b
2) such an element is 0
3) this is true: set z=a
4) this is blindingly obviously true.
The set "{1,2,3,4} with the operation *" is a group, where in the following table, the entry in the mth row and the nth column gives m*n:
1234
2143
3412
4321
(it's clear from the table that the first axiom holds; 1 is our e; each element a has a*a=e; and it can be tediously checked that the fourth axiom holds.)
Groups do not have to be "commutative"  that is, a+b isn't necessarily b+a. However, the first example is of size 6, and it's a pain (and not particularly helpful) to write out the table.
A subgroup H of G is a subset of G such that H is a group. Hence, of course, it must contain the same identity.
Prove Lagrange's Theorem: if H is a subgroup of G, then H divides G.
Hint:
Spoiler:Show
Hence prove that if g is in G, then the order of g (that is, the minimum n such that g^n is the identity) divides G.
Corollary: Prove Fermat's Little Theorem: that if then .Last edited by Smaug123; 28072013 at 22:13. Reason: Added groups summary 
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 2020
 28072013 22:39
(Original post by FireGarden)
You can; they are the basis vectors of functions used when dealing with fourier series.
They are orthogonal under the innerproduct defined by
edit: Problem 291*
Prove the above statement that sines and cosines are orthogonal under the inner product given. This is to show the integral of , , or , are zero if and only if
As, if, , we have,
Solution 291
Let, m be distinct from n. Then integrating by parts twice, we have,
But, as .
You get exactly the same result when you perform .
Further, if , we simply have,
And that should be enough to imply that the inner product is 0 iff .
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