Conference rusure::math

>    Which of the two integrals
>                
>    	Sn = int(int(...int(sin(x1+x2+...+xn),
>    		x1=0..pi/2),x2=0..pi/2)...,xn=0..pi/2)
>    
>
>    	Cn = int(int(...int(cos(x1+x2+...+xn),
>    		x1=0..pi/2),x2=0..pi/2)...,xn=0..pi/2)
>                
>    is larger?
    
From trig formulae:

	(Sn)   ( 1 1) (S(n-1))
	(Cn) = (-1 1)*(C(n-1))

Find the eigenvalues
Express Sn & Cn as sums of powers of eigenvalues
Use boundary conditions S0 = 0, C0 = C1 = S1 = 1 to derive constants
Next spot the simplification:

	Sn = 2^(n/2)*sin(n*pi/4)
	Cn = 2^(n/2)*cos(n*pi/4)

Andrew

    Solution by Kevin Ford (student), University of Illinois at
    Urbana-Champaign, IL.
    
    We claim that Sn = Cn when n = 1, 5 (mod 8), Sn > Cn when n = 2, 3, 4
    (mod 8), and Sn < Cn when n = 0, 6, 7 (mod 8).  Since
    
    	Cn + i*Sn = int(int(...int(e^(i*(x1+x2+...xn)),x1=0..pi/2),
    		x2=0..pi/2)...,xn=0..pi/2)
    		= int(e^(i*x),x=0..pi/2)^2
    		= (1+i)^n
    
    we have arg(C+i*Sn) = pi*n/4, and the claim follows.

Kevin lost me on his first line:

  	Cn + i*Sn = int(int(...int(e^(i*(x1+x2+...xn)),x1=0..pi/2)


I'm most puzzled by how the two nested integrals got combined into just one.

/Eric

re .3
    
>Kevin lost me on his first line:
>
>  	Cn + i*Sn = int(int(...int(e^(i*(x1+x2+...xn)),x1=0..pi/2)
>
>I'm most puzzled by how the two nested integrals got combined into just one.
    
    Not sure whether you mean that you can't see that the first line
    (reproduced above) is correct or how he got from the first line to the
    second (in which I would point out *n* nested integrals got combined
    into just one).
    
    Personally I found the published proof a little short on working for
    mere mortals such as myself (but perhaps suitable for the forum in
    which it appeared).
    
    I think in any case that there is a typo viz.
    
>    		= int(e^(i*x),x=0..pi/2)^2
    
    should read
    
    		= int(e^(i*x),x=0..pi/2)^n
    
    Here's more detailed working.
    
    	int(int(...int(e^(i*(x1+x2+...xn)),x1=0..pi/2),
    		x2=0..pi/2)...,xn=0..pi/2)
    
= int(e^(i*x1),x1=0..pi/2)�int(e^(i*x2),x2=0..pi/2�...int(e^(i*xn),xn=0..pi/2)
    
    All that has been done here is to distribute the multiplication over
    the addition and then use e^(a+b) = e^a�e^b and then recognise that the
    integrals can be split apart.
    
    But now each of the n integrals multiplied together is identical so he
    introduces a new variable to do one integration with and raises the
    result to the nth power
    
= int(e^(i*x),x=0..pi/2)^n
    
    Now evaluating the integral...
    
      int(e^(i*x),x=0..pi/2)
    
      1          pi/2
    = - [e^(i*x)]
      i          0
    
      1
    = - (i - 1)
      i
    
      (i - 1)i 
    = --------
         i�
    
      -1-i
    = ----
       -1
    
    = 1+i
    
    Hence the original integral is (1+i)^n
    
    Does that clarify adequately?

    Re .4:
    
    Yes, the exponent should be n, not 2.
    
    
    Re .3:
    
    To combine the two integrals, just multiply Sn by i (a constant times
    the integral of f(x) equals the integral of the constant times f(x))
    and add it to Cn (the integral of f(x) plus the integral of g(x) equals
    the integral of f(x) plus g(x)).  Then you have cos(x...)+i*sin(x...)
    inside the integrals, which is equal to e^(i*(...)).
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

>    	int(int(...int(e^(i*(x1+x2+...xn)),x1=0..pi/2),
>    		x2=0..pi/2)...,xn=0..pi/2)
>    
>= int(e^(i*x1),x1=0..pi/2)�int(e^(i*x2),x2=0..pi/2�...int(e^(i*xn),xn=0..pi/2)
>    
>    All that has been done here is to distribute the multiplication over
>    the addition and then use e^(a+b) = e^a�e^b and then recognise that the
>    integrals can be split apart.


I guess i'm just getting rusty with this stuff.  It's been awhile.  Your
details were quite clear, but I'm a bit worried about this bit about "integrals
can be split apart".

In general, if we have

	int(int(f1(x)*f2(x)))

can we really say it's equivalent to

	int(f1(x))*int(f2(x))

I just don't remember this...

/Eric

    Re .6:
    
    In general, int(f(x)*g(x),x) is NOT equal to int(f(x),x)*int(g(x),x).
    
    However, int(int(f(x0)*g(x1),x0),x1) is equal to
    int(int(f(x0),x0)*g(x1),x1).  That is, g(x1) is constant with respect
    to x0, so you can move it outside of the x0 integral.
    
    
    				-- edp
    
    
Public key fingerprint:  8e ad 63 61 ba 0c 26 86  32 0a 7d 28 db e7 6f 75
To find PGP, read note 2688.4 in Humane::IBMPC_Shareware.

additional to .7                                                    
    
>    However, int(int(f(x0)*g(x1),x0),x1) is equal to
>    int(int(f(x0),x0)*g(x1),x1).  That is, g(x1) is constant with respect
>    to x0, so you can move it outside of the x0 integral.
    
    And then going further, int(f(x0),x0) is constant with respect to x1,
    so you can move it outside of the x1 integral. Hence the expression can
    be written
    
    int(f(x0),x0) * int(g(x1),x1)
    
re .0
    
    It occurs to me that the published solution should probably indicate
    that n is a *strictly positive* integer i.e. the case n = 0 isn't
    really well defined and yet is included in the claimed solution.