Jim requested to post the following question:
1. What is the size or cardinality of L1[a,b], the space of the Lebesgue integrable functions on [a,b]
Also,
2. Find a polynomial f(x) with real coefficients and f(0) = 1, such that the sums of the squares of the coefficients of f(x)n and (3x2 + 7x + 2)n are the same.
3. Let C be the circle radius 1, center the origin. A point P is chosen at random on the circumference of C, and another point Q is chosen at random in the interior of C. What is the probability that the rectangle with diagonal PQ, and sides parallel to the x-axis and y-axis, lies entirely inside (or on) C?
4. G is a group consisting of m real n x n matrices with the operation of matrix multiplication. The sum of the traces of the elements of G is zero. Prove that the sum of the elements of G is the zero matrix.
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Solution for problem 2
ReplyDeleteThe good news is that I have a solution for the problem, and the bad news is that my proof is not very elegant. The solution that work is: 6x²+5x+1.We start by computing the sum of the squares of coefficient of 3x²+7x+2, which is: 3²+49+4= 62. The polynomial that we are looking for must have for constant term 1. Thus, the sum of the squares of the other terms must be 62-1=61. So we look for perfect squares that will add up to 61. Focusing on sums of perfect squares that add to a number that ends up with just 1, we get 5 and 6. In fact 5²+6²= 61. So, the coefficients of f(x) must follow the pattern 5,6,1 or 6,5,1. I then check for the first pattern: f(x)=5x²+6x+1. This fails because the sum of the squares of coefficient of (3x²+7x+2)²=28x+61x²+42x³+9x⁴+4 is 6366 which doesn't match up with the sum of the squares of coefficients of (5x²+6x+1)²= 12x+46x²+60x³+25x⁴+1. Since the pattern 5,6,1 fails, we attempt the other possible pattern which is 6,5,1. Therefore I tried for f(x)=6x²+5x+1 and I obtained a perfect match. I checked up to n=4, and obtained for the sums of perfect squares: 62, 6366, 751520, and 93950470. Thus the answer is indeed confirmed to be 6x²+5x+1.
Vignon
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ReplyDeleteOne remark is that m=2 doesn't work for 2 by 2 matrices. However, I was able to check that the claim works for a group of 2 by 2 matrices of order 3. Here is a group of order 3 that works for some real number b,
ReplyDeleteG={[1,0;0,1],[-2,b;-3/b,1],[1,-b;3/b,-2]}. So this problem is only true for some particular m. I want to attempt to generalize this proof for any m that works but I have a feeling that there is some easier way to prove this.
Vignon