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    (Original post by Smaug123)
    I think R^n is fine for "nD space" - although I would probably say "it is the length of the projection onto the span of x1,x2,x3" - not sure, though.
    Obviously \mathbb{R}^2 is the 2-D Euclidean plane, but I'm not sure whether that is the same as an arbitrary plane.
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    (Original post by henpen)
    Obviously \mathbb{R}^2 is the 2-D Euclidean plane, but I'm not sure whether that is the same as an arbitrary plane.
    It is - "the plane spanned by these two vectors" is isomorphic to R^2, anyway.
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    (Original post by FireGarden)
    You can; they are the basis vectors of functions used when dealing with fourier series.

    They are orthogonal under the inner-product defined by  <f,g> = \displaystyle\int_{-\pi}^{\pi}f(x)g(x)\ dx

    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 \sin(mx)\cos(nx), \cos(mx)\cos(nx), or \sin(mx)\sin(nx), are zero if and only if  m \not= n
    I came across this really nice derivation on MIT OCW DE:

    Let u,v be functions of x that satisfy
    \displaystyle u''=-m^2u, v''=-n^2v

    \displaystyle \int_{-\pi}^{\pi}u''vdx=[u'v]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}u'v'dx
    \displaystyle \int_{-\pi}^{\pi}v''udx=[v'u]_{-\pi}^{\pi}-\int_{-\pi}^{\pi}v'u'dx

    Note that [u'v]_{-\pi}^{\pi}=0 for all m,n.
    Next,

    \displaystyle \int_{-\pi}^{\pi}u''vdx= -m^2\int_{-\pi}^{\pi}uvdx=-\int_{-\pi}^{\pi}u'v'dx
    \displaystyle -n^2\int_{-\pi}^{\pi}vudx=-\int_{-\pi}^{\pi}v'u'dx

    Thus
    \displaystyle -m^2\int_{-\pi}^{\pi}uvdx=-n^2\int_{-\pi}^{\pi}vudx

    Obviously if m=n the equality holds for all solutions u,v, but if m,n are not equal the equality holds only if the integral is equal to 0.
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    (Original post by Jkn)
    Yeah that;s what I had in mind! "inner product" I think it's called?

    || \mathbf{v} || = \sqrt{\sum_{i=1}^{n} v_i^2}

    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?

    It's a specific case of the inner product (when v multiplies itself), I believe the inner product of two vectors is a generalisation of the dot product (allowing functions, numbers and vectors to multiply together using a single 'function', although at the moment it seems a bit of a useless generalisation, just giving the same name to different things).
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    (Original post by henpen)
    It's a specific case of the inner product (when v multiplies itself), I believe the inner product of two vectors is a generalisation of the dot product (allowing functions, numbers and vectors to multiply together using a single 'function', although at the moment it seems a bit of a useless generalisation, just giving the same name to different things).
    The inner product is a very general thing that is a property of all vector spaces - they're not all just R^n. It provides a measure of "distance" between two points (those points may themselves represent functions, numbers, vectors, etc).
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    (Original post by Jkn)
    Yeah that;s what I had in mind! "inner product" I think it's called?
    The inner product is actually a gorgeous concept. It needn't be constrained to this. In fact, you can define your own inner product, (so long as it satisfies a set of the fundamental axioms) upon a set of elements and thereby construct your own inner product space.
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    Here's a fun one, offering quite a few interesting solutions:

    Problem 292

    If  n \in \mathbb{Z} and  p is a prime such that  p| (n^4 - n^2 +1 ) , show that  p is of the form  12k+1 .
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    (Original post by MW24595)
    The inner product is actually a gorgeous concept. It needn't be constrained to this. In fact, you can define your own inner product, (so long as it satisfies a set of the fundamental axioms) upon a set of elements and thereby construct your own inner product space.
    If it satisfies said axioms, what properties does it have (basically, why is it useful)? What are these axioms?
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    (Original post by henpen)
    If it satisfies said axioms, what properties does it have (basically, why is it useful)? What are these axioms?
    Here they are:

    (i) <v, v> ≥ 0 for all v \in V,
    (ii) <v, v> = 0 if and only if  v = 0,
    (iii) <αv,w> = αv,w for all  α \in R and all  v,w \in V,
    (iv) <u, v + w >= <u, v> + <u,w> for all  u, v,w \in V , and finally,
    (v) <v,w> = <w, v> for all v,w \in V.


    Coming up with such binary relations can lead to all sorts of interesting structures, atleast from a group-theoretic perspective. :rolleyes:

    On that note, it is incredible just how much we take commutativity for granted. When one deals with general abstract nonsense, one can't help but pray for commutativity. The theory is just too pretty.
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    (Original post by MW24595)
    Here's a fun one, offering quite a few interesting solutions:

    Problem 292

    If  n \in \mathbb{Z} and  p is a prime such that  p| (n^4 - n^2 +1 ) , show that  p is of the form  12k+1 .
    Solution 292:

     n^4-n^2+1 = (n^2-\frac{1}{2})^2+\frac{3}{4}

    This quadratic has no real roots and therefore has no real factorisations except for:

     1(n^4-n^2+1)

    As such, it is now sufficient to prove that  n^4-n^2 is divisible by 12. Rewrite it as:

     n^2 (n^2-1)

     n^2(n-1)(n+1)

    Now, if n is odd, you get two factors of 2 from each of the linear terms. If n is divisible by 3, then a factor of 3 occurs in the n^2. If not, it must occur in n-1 or n+1 as these are three consecutive integers, so divisible by 12.

    If n is even, then you get two factors of two from n^2 and a factor of three from n, n-1 or n+1, and so it is divisible by, 12 as required.

    Hence any prime factor of the given equation is in the form 12k+1, as required.
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    (Original post by MW24595)
    On that note, it is incredible just how much we take commutativity for granted. When one deals with general abstract nonsense, one can't help but pray for commutativity. The theory is just too pretty.
    Seconded - one of my favourite theorems is "If a group has order p or p^2 for prime p, then it is abelian". [Abelian = commutative, in group theory.] That and Sylow, anyway
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    Solution 292

    We have (n^{2}-1)^{2} \equiv -n^{2} \pmod p; hence p \equiv 1 \pmod 4.
    Also, n^{4}+2n^{2}+1-3n^{2} \equiv 0 \pmod p, so p \equiv 1 \pmod 3, since the case p=3 is not possible.
    Hence the result.

    Solution 261

    The first part is trivial. Consider the second.
    It is clear that (p-2)! \equiv 1 \pmod p, giving \displaystyle \left(\frac{2}{p} \right) =1 - contradiction.
    Now 2(p-3)! \equiv -1 \pmod p, and thus 2m^{2} \equiv 1 \pmod p, or \displaystyle \left(\frac{2}{p} \right) =1, which is a contradiction once again.
    Therefore, no solutions of the forms (p-2,m) and (p-3,m) exist.
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    (Original post by Smaug123)
    Seconded - one of my favourite theorems is "If a group has order p or p^2 for prime p, then it is abelian". [Abelian = commutative, in group theory.] That and Sylow, anyway
    Here's another one:

    If a group has order  pq , where  p \not \equiv 1 mod q or vice versa, then the group is abelian.
    Furthermore, it is cyclic!

    This actually follows from the result I proved on the ASOM thread from your homework.
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    (Original post by DJMayes)
    Solution 292:
    .
    I got to the same point using quadratic residues, but why does it follow from n^4-n^2+1 being of the form 12k+1 that all its prime factors must be? Could it not have other prime factors?

    You're correct DJMayes, I didn't notice the importance of the no real roots part of the proof.
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    (Original post by henpen)
    I got to the same point using quadratic residues, but why does it follow from n^4-n^2+1 being of the form 12k+1 that all its prime factors must be? Could it not have other prime factors?
    My argument is that the only way the quadratic can be factorised into integers is as 1 multiplied by itself, as the quadratic in question cannot be factorised into real linear roots. As p=1 is not the factor, p must then be the function itself, from which my argument follows. If I have missed out some intricacies of the factorisation please correct me.
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    Solution 292

    The quadratic residues of both 4 and 3 are {0,1}. Thus
    \displaystyle n^2(n^2-1)+1 \equiv 1 \mod 3,4
    \displaystyle n^2(n^2-1)+1 \equiv 1 \mod 12
    \displaystyle n^2(n^2-1)+1 = 12k+1

    Thus some prime factors of n^2(n^2-1)+1 are of the form  12k+1. Assume there are prime factors not of that form, then they will multiply with the previous prime factors to make a number of the form  12kp+p, where p is a number other than 1, contradicting the claim that  n^2(n^2-1)+1 = 12k+1.

    I find that a little more intuitive.
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    Solution 292

    Here's another one. Forgive me, for I skipped some working, I shall add the minutiae in later.
    It also involves quite a few intermediary results, but, ah, well.

    If  p| n^4 - n^2 + 1 \Rightarrow (-1)^2 - 4(1)(1)= -3 is a quadratic residue mod p. Let  p' = \frac{p-1}{2} .

    So, via Gauss' Lemma, we have  \left( \frac{-3}{p} \right) = (-1)^S , where,  S = \displaystyle \sum_{k=1}^{p'} \lfloor  \frac{-6k}{p} \rfloor  .

    Now, we're primarily concerned with  S = \displaystyle \sum_{k=1}^{p'} \left \lfloor  \frac{-6k}{p} \right \rfloor \ mod \ 2 , because that's the only deciding factor when  -1 is raised to that exponent.

    And so,

     \displaystyle \sum_{k=1}^{p'} \left \lfloor  \frac{-6k}{p} \right \rfloor = p' - \left \lfloor \frac{2}{3} p' \right \rfloor + \left \lfloor \frac{1}{3} p' \right \rfloor = \left \lfloor \frac{p+1}{6} \right \rfloor + \left \lfloor \frac{p-1}{6} \right \rfloor

    Now, in order for this sum to be even, and therefore for  -3 to be a quadratic residue,  p \not \equiv 5 \ mod \ 6 \Rightarrow p \equiv 1 \ mod \ 6.

    On the other hand,  n^4 - n^2+1 = (n^2-1)^2 + n^2 , and further,  gcd((n^2-1)^2,  n^2) = 1 . So,  p \equiv 1 \ mod \ 4 .

    Combining these 2 results using the Chinese Remainder Theorem, we can say that,   p \equiv 1 \ mod \ 12 .
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    Problem 293

    Prove that

    \displaystyle \int_1^{\infty} \ln \left( \frac{\ln(x)}{x} \right) \left(\frac{ \ln(x)}{x} \right)^ndx = \frac{\Gamma(n+1)}{(n-1)^{n+1}}\left(\psi(n+1)-\frac{n+1}{n-1} - \ln(n-1) \right)

    Sorry if it's a bit mundane.
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    (Original post by henpen)
    Problem 293

    Prove that

    \displaystyle \int_1^{\infty} \ln \left( \frac{\ln(x)}{x} \right) \left(\frac{ \ln(x)}{x} \right)^ndx = \frac{\Gamma(n+1)}{(n-1)^{n+1}}\left(\psi(n+1)-\frac{n+1}{n-1} - \ln(n-1) \right)

    Sorry if it's a bit mundane.
    Finally an integral..

    Solution 293

    Let x \mapsto e^{\frac{x}{n-1}},

    \displaystyle \int_1^{\infty} \ln \left( \frac{\ln(x)}{x} \right) \left(\frac{ \ln(x)}{x} \right)^n \ dx = \frac{1}{(n-1)^{n+1}} \int_0^{\infty} \left( \ln(x)-\ln(n-1)-\frac{x}{n-1} \right) x^n e^{-x} \ dx
    \displaystyle = \frac{1}{(n-1)^{n+1}} \left(\frac{d}{dn} \Gamma (n+1) - \ln(n-1) \Gamma(n+1) - \frac{\Gamma (n+2)}{n-1} \right)
    \displaystyle = \frac{\Gamma(n+1)}{(n-1)^{n+1}}\left(\psi(n+1)-\frac{n+1}{n-1} - \ln(n-1) \right) as required.
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    (Original post by Jkn)
    Finally an integral..

    Solution 293

    Let x \mapsto e^{\frac{x}{n-1}},

    \displaystyle \int_1^{\infty} \ln \left( \frac{\ln(x)}{x} \right) \left(\frac{ \ln(x)}{x} \right)^n \ dx = \frac{1}{(n-1)^{n+1}} \int_0^{\infty} \left( \ln(x)-\ln(n-1)-\frac{x}{n-1} \right) x^n e^{-x} \ dx
    \displaystyle = \frac{1}{(n-1)^{n+1}} \left(\frac{d}{dn} \Gamma (n+1) - \ln(n-1) \Gamma(n+1) - \frac{\Gamma (n+2)}{n-1} \right)
    \displaystyle = \frac{\Gamma(n+1)}{(n-1)^{n+1}}\left(\psi(n+1)-\frac{n+1}{n-1} - \ln(n-1) \right) as required.
    Pretty much. A more direct way would be to differentiate the answer to problem 22, which of course is implicitly what you did.
 
 
 
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