Hey there! Sign in to join this conversationNew here? Join for free
    Offline

    2
    ReputationRep:
    (Original post by henpen)
    Solution 286
    For even n,

    \displaystyle \prod_{m=1}^{2r}\Gamma \left(\frac{m}{2r+1} \right)=\displaystyle \prod_{m=1}^{r}\frac{\pi}{\sin \left(\frac{\pi m}{2r+1} \right)} \right)= \frac{\pi^r 2^{r}}{\sqrt{2r+1}}

    For odd n, I'll try to generalise 24 later, but I need to sleep now.
    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!
    Hence, we have: \displaystyle \prod_{k=1}^{m-1} \Gamma \left( \frac{k}{m} \right) = \frac{(2 \pi)^{\frac{m-1}{2}}}{\sqrt{m}} which implies that the answer was \frac{(2 \pi)^{\frac{9}{2}}}{\sqrt{10}}
    which I suppose can be seen as yet another reason to introduce \tau as being equal to 2 \pi.

    Does anyone have any strong views on this? Is \pi more arbitrary than \tau?

    This is essentially a consequence of the multiplication theorem in the case where we let z=\frac{1}{m}.
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by Jkn)
    Does anyone have any strong views on this? Is \pi more arbitrary than \tau?
    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).
    Offline

    2
    ReputationRep:
    (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).
    '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).

    Problem 290
    *

    Evaluate \displaystyle \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma in the case where \alpha, \beta, \gamma are, in 3-dimensional 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 n-dimensional Euclidean space where the co-ordinates lie in \mathbb{R}^n? (bonus-only ***)
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (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).
    They're both so un-arbitrary that the choice between them is arbitrary

    (quibble with the wording of your Problem 290: "any three orthonormal bases" should be "any three orthonormal vectors" or "any orthonormal basis")
    Offline

    2
    ReputationRep:
    (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")
    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?
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (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?
    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 some
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (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 x^3+ax^2=y^2. Given that  1 \leq x \leq b where b is a positive integer, find, in terms of a and b, the number of possible pairs (x,y) that satisfy the equation.
    Is this not just \lfloor \sqrt{a+b} \rfloor? Since the expression is x^2 (a+x) = y^2, 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
    Offline

    0
    ReputationRep:
    (Original post by Jkn)
    Well done man, I'll say that was blagged!
    Sort of, I can't seem to generalise result 24 when the '1006' is odd: the argument of \prod_{k=1}^{2r-1}e^{\frac{ik2 \pi}{2r}} isn't equal to 0, so the result isn't as simple.

    Wait, I may have gotten that wrong earlier in my sleep-induced stupor, it may very well generalise.
    Offline

    0
    ReputationRep:
    (Original post by Jkn)

    Problem 290
    *

    Evaluate \displaystyle \cos^2 \alpha + \cos^2 \beta + \cos^2 \gamma in the case where \alpha, \beta, \gamma are, in 3-dimensional 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 n-dimensional Euclidean space where the co-ordinates lie in \mathbb{R}^n? (bonus-only ***)
    'Solution' 290
    alp

    Take a vector such that |\mathbf{v}|=1. \cos(\alpha_1) is its projection onto the x_1-axis, \cos(\alpha_2) onto the x_2-axis ... .

    Thus \sqrt{\cos^2( \alpha_1)+\cos^2( \alpha_2)} is its projection onto the x_1x_2-plane, and \sqrt{\cos^2( \alpha_1)+\cos^2(\alpha_2)+\cos^  2(\alpha_3)} is its projection onto the x_1x_2x_3-'3D space' (what word do I use here, \mathbb{R}^3?), and \sqrt{\sum_{k=1}^n\cos^2( \alpha_n)} is its projection onto the x_1x_2...x_n-'nD space'. But this projection is a line of length 1, so \sqrt{\sum_{k=1}^n\cos^2( \alpha_n)}=1 and \sum_{k=1}^n\cos^2( \alpha_n)=1.

    I hope you understood what I meant. Could you help formalise the bits where I was hopelessly off with regards to the mathematical language?
    Offline

    0
    ReputationRep:
    (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 some
    Can you not also have \sin(kx),\cos(kx), k \in \mathbb{N} as orthonormal basis vectors? I hear that definition's got something to with them being orthogonal under integration (between certain limits).
    Offline

    2
    ReputationRep:
    (Original post by Smaug123)
    Is this not just \lfloor \sqrt{a+b} \rfloor? Since the expression is x^2 (a+x) = y^2, 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
    Looks right! I think people just tried to over-complicate things and so got put off :lol:

    Why not try some of my fiendish integral/E-M/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 \prod_{k=1}^{2r-1}e^{\frac{ik2 \pi}{2r}} isn't equal to 0, so the result isn't as simple.

    Wait, I may have gotten that wrong earlier in my sleep-induced stupor, it may very well generalise.
    Why not try a different approach?

    Actually, what I found most trilling is that we now have a rather strange alternate solution to problem 24 using the ever-powerful gamma function (i.e. multiplication theorem). Waaaay too hungover to type it up though.. :|
    Offline

    2
    ReputationRep:
    (Original post by henpen)
    'Solution' 290
    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?
    I hope you understood what I meant. Could you help formalise the bits where I was hopelessly off with regards to the mathematical language?
    Smaug seemed far more well-versed with vector terminology etc.. so I will leave it to him to formalize your language (seems fine to me :dontknow:).
    Offline

    0
    ReputationRep:
    (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 well-versed with vector terminology etc.. so I will leave it to him to formalize your language (seems fine to me :dontknow:).
    Okay, how about: choose the vector \mathbf{v} such that |\mathbf{v}|=1 and let \mathbf{x}_k,k \in \{1,...,n\} be the orthonormal basis vectors in \mathbb{R}^n.

    Expand the vector componentwise,:
    \displaystyle \mathbf{v}= \sum_{i=1}^n \mathbf{x}_i v_i

    and its modulus:
    \displaystyle |\mathbf{v}|^2=1= \sum_{i=1}^n v_i^2

    But note that \left \langle \mathbf{v},\mathbf{x}_i  \right \rangle =v_i by the vector's expansion, and \displaystyle\left \langle \mathbf{v},\mathbf{x}_i  \right \rangle = |\mathbf{v}||\mathbf{x}_i|\cos( \alpha _{\mathbf{x}_i, \mathbf{v}}) by the definition of  \alpha , and thus v_i= \cos( \alpha _{\mathbf{x}_i, \mathbf{v}}), giving

    \displaystyle |\mathbf{v}|^2=1= \sum_{i=1}^n \cos^2( \alpha _{\mathbf{x}_i, \mathbf{v}})
    Offline

    3
    ReputationRep:
    (Original post by henpen)
    Can you not also have \sin(kx),\cos(kx), k \in \mathbb{N} as orthonormal basis vectors? I hear that definition's got something to with them being orthogonal under integration (between certain limits).
    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
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by henpen)
    'Solution' 290
    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.
    Offline

    2
    ReputationRep:
    (Original post by henpen)
    and its modulus:
    \displaystyle |\mathbf{v}|^2=1= \sum_{i=1}^n v_i^2
    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?
    Offline

    2
    ReputationRep:
    (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
    I remember that from QM! (one function should be conjugated btw, unless their codomain is real)
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    (Original post by Jkn)
    I remember that from QM! (one function should be conjugated btw, unless their codomain is real)
    Even if their codomain is real, it doesn't hurt
    • PS Helper
    • Study Helper
    Offline

    13
    ReputationRep:
    PS Helper
    Study Helper
    Problem 292***
    Group summary:
    Spoiler:
    Show
    I 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
    Consider the cosets of H in G - that is, for a given a \in G, consider the set \{ah, h \in H \} . How many of these sets are there as we vary a? What size are they?

    Hence prove that if g is in G, then o(g) 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 (a,p) = 1 then a^{p-1} \equiv 1 \pmod{p}.
    Offline

    0
    ReputationRep:
    (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 think you mean,  |m| \not= |n| .

    As, if,  m = -n , we have,  <\sin {mx}, \sin {nx}> = - \frac{\pi}{m}.

    Solution 291

    Let, m be distinct from n. Then integrating by parts twice, we have,

     \displaystyle \int_{-\pi}^{\pi} \sin {mx} \sin {nx}\ dx = \frac{n}{m} \displaystyle \int_{-\pi}^{\pi} \cos {mx} \cos {nx}\ dx = \frac{n^2}{m^2} \displaystyle \int_{-\pi}^{\pi} \sin {mx} \sin {nx}\ dx

    But, as  m^2 \not= n^2 \Rightarrow \displaystyle \int_{-\pi}^{\pi} \sin {mx} \sin {nx}\ dx = 0.

    You get exactly the same result when you perform  \displaystyle \int_{-\pi}^{\pi} \cos {mx} \cos{nx}\ dx .

    Further, if  m=n , we simply have,

     \displaystyle \int_{-\pi}^{\pi} \sin^2 {mx} dx =  \displaystyle \int_{-\pi}^{\pi} \cos^2 {mx} dx= \frac{1}{2m}\displaystyle \int_{-\pi}^{\pi}  1- \cos{2x}\ dx= \frac{ \pi}{m}

    And that should be enough to imply that the inner product is 0 iff  |m| \not= |n| .
 
 
 
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • Poll
    Would you like to hibernate through the winter months?
    Useful resources

    Make your revision easier

    Maths

    Maths Forum posting guidelines

    Not sure where to post? Read the updated guidelines here

    Equations

    How to use LaTex

    Writing equations the easy way

    Student revising

    Study habits of A* students

    Top tips from students who have already aced their exams

    Study Planner

    Create your own Study Planner

    Never miss a deadline again

    Polling station sign

    Thinking about a maths degree?

    Chat with other maths applicants

    Can you help? Study help unanswered threads

    Groups associated with this forum:

    View associated groups
  • See more of what you like on The Student Room

    You can personalise what you see on TSR. Tell us a little about yourself to get started.

  • The Student Room, Get Revising and Marked by Teachers are trading names of The Student Room Group Ltd.

    Register Number: 04666380 (England and Wales), VAT No. 806 8067 22 Registered Office: International House, Queens Road, Brighton, BN1 3XE

    Quick reply
    Reputation gems: You get these gems as you gain rep from other members for making good contributions and giving helpful advice.