document.write( "Question 452779: Prove that for any n >= 2 , given .\r
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Algebra.Com's Answer #311486 by Edwin McCravy(20056)\"\" \"About 
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document.write( "Prove: for \"n%3E=2\"\r\n" );
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document.write( "Euler's equation: \"e%5E%28i%2Atheta%29+=+cis%28theta%29+=+cos%28theta%29+%2B+i%2Asin%28theta%29\"\r\n" );
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document.write( "From which we can get: \r\n" );
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document.write( "\"sin%28theta%29+=+%28e%5E%28i%2Atheta%29+-+e%5E%28-i%2Atheta%29%29%2F%282i%29\"\r\n" );
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document.write( "So each of the factors in the original expression is:\r\n" );
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document.write( "where k goes from 1 to n-1\r\n" );
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document.write( "Factor out the first term:\r\n" );
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document.write( "The original expression becomes a product of these three things:\r\n" );
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document.write( "#1.  \r\n" );
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document.write( "#2. \r\n" );
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document.write( "#3. \r\n" );
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document.write( "where k goes from 1 to n-1\r\n" );
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document.write( "Adding the exponents in #1.\r\n" );
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document.write( "#1.  \r\n" );
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document.write( "The sum of the first n-1 integers is \"%28n-1%29n%2F2\", so the\r\n" );
document.write( "exponent of e becomes\r\n" );
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document.write( "\"%28i%2Api%2A%28n-1%29n%2F2%29%2Fn=i%2Api%2A%28n-1%29n%2F%282n%29=i%2Api%2A%28n-1%29%2F2+\"\r\n" );
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document.write( "and #1 is now \r\n" );
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document.write( "#1. \r\n" );
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document.write( "Since #3  \r\n" );
document.write( "it gives the product:\r\n" );
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document.write( "\"%281%2F%282i%29%29%5E%28n-1%29+=+%28-i%2F2%29%5E%28n-1%29+\"\r\n" );
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document.write( "Putting #1 and #3 together:\r\n" );
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document.write( "#1*#3  \"i%5E%28n-1%29%28-i%2F2%29%5E%28n-1%29=%28-i%5E2%2F2%29%5E%28n-1%29=1%2F%282%5E%28n-1%29%29\"\r\n" );
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document.write( "Now we look at #2 again.\r\n" );
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document.write( "#2. \r\n" );
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document.write( "Those factors are all (1 - an nth root of 1 other than 1 itself. To show that\r\n" );
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document.write( "1 = cis(0 + 2k*pi) = e^(2k*pi) \r\n" );
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document.write( "and by DeMoivre's theorem the n nth roots of unity are \r\n" );
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document.write( "cis(2k*pi/n) and\r\n" );
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document.write( "that's what the second terms in those parenthetical expressions are.\r\n" );
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document.write( "and we can use any n-1 consecutive even integers for k, so here we are using\r\n" );
document.write( "the n-1 negative consecutive even integers -2, -4, -6,  -2(n-1). And they\r\n" );
document.write( "won't include 1 itself because we are not including integers 0 or 2n.\r\n" );
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document.write( "The nth roots of 1 can also be gotten by solving the equation\r\n" );
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document.write( "\"x%5En+=+1\"\r\n" );
document.write( "\"x%5En+-+1+=+0\"\r\n" );
document.write( "\"%28x-1%29%28x%5E%28n-1%29%2Bx%5E%28n-2%29%2Bx%5E%28n-3%29%2B%22...%22%2B1%29+=+0\"\r\n" );
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document.write( "The first parentheses give us the root of one which is 1 itself.\r\n" );
document.write( "The n-1 degree polynomial in the second parentheses has solutions\r\n" );
document.write( "which are all the n-1 nth roots of 1 other than 1.\r\n" );
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document.write( "Therefore the polynomial\r\n" );
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document.write( "#4. x^(n-1)+x^(n-2)+x^(n-3)+\"...\"+1 \r\n" );
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document.write( "is equivalent to a polynomial which is like #2 with\r\n" );
document.write( "x's placed where the 1's are.\r\n" );
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document.write( "#5. \r\n" );
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document.write( "because they have the same solution and both have leading\r\n" );
document.write( "coefficient 1.\r\n" );
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document.write( "So when we substitute 1 for x in #5, we get #2 and when\r\n" );
document.write( "we substitute 1 for x in #4 we get n because there are n terms.\r\n" );
document.write( "Therefore #2 simplifies to just n\r\n" );
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document.write( "Therefore #1*#2*#3 = \"n%2F2%5E%28n-1%29\"\r\n" );
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document.write( "Edwin

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