Constructibility and the construction of a 17-sided
... is the length of each side. This gives a solution of s = 2, a constructible real number. Therefore we are able to construct a square with an area of 4 units2 . We can apply this concept to more complicated algebraic equations. There exist unique equations to represent lines and circles. A line can b ...
... is the length of each side. This gives a solution of s = 2, a constructible real number. Therefore we are able to construct a square with an area of 4 units2 . We can apply this concept to more complicated algebraic equations. There exist unique equations to represent lines and circles. A line can b ...
Group-theoretic algorithms for matrix multiplication
... In fact, the reader familiar with Strassen’s 1987 paper [10] and Coppersmith and Winograd’s paper [3] (or the presentation of this material in, for example, [1]) will recognize that our exponent bounds of 2.48 and 2.41 match bounds derived in those works. It turns out that with some effort the algor ...
... In fact, the reader familiar with Strassen’s 1987 paper [10] and Coppersmith and Winograd’s paper [3] (or the presentation of this material in, for example, [1]) will recognize that our exponent bounds of 2.48 and 2.41 match bounds derived in those works. It turns out that with some effort the algor ...
32(2)
... m = 3: The assertion P3 ID /}; P2 is equivalent to x3 DX;XJ. m = 4; Note that x4 = aif^, x = d^, and x3 = i^. Therefore, applying concatenation to the alignments cdz)d;c and P4 ZDP2;P3 implies that x4 IDX;CX3. Consequently, by Lemma 1,(1) cannot hold for m = 4, since x2 begins with a rf. Similar rea ...
... m = 3: The assertion P3 ID /}; P2 is equivalent to x3 DX;XJ. m = 4; Note that x4 = aif^, x = d^, and x3 = i^. Therefore, applying concatenation to the alignments cdz)d;c and P4 ZDP2;P3 implies that x4 IDX;CX3. Consequently, by Lemma 1,(1) cannot hold for m = 4, since x2 begins with a rf. Similar rea ...
Full text
... Fn−1 Fn+1 = Fn2 + (−1)n , it follows that (Fn−1 Fn ) ⊕ (Fn Fn+1 ) = (Fn−1 + Fn )(Fn + Fn+1 ) = Fn+1 Fn+2 . and so the sequence xn := Fn Fn+1 satisfies the modified Fibonacci recurrence xn+1 = xn ⊕ xn−1 . Here we define the first new sequence. Definition 3.4. Let b1 = 0, and for n ≥ 1, b2n = bn b2n+1 ...
... Fn−1 Fn+1 = Fn2 + (−1)n , it follows that (Fn−1 Fn ) ⊕ (Fn Fn+1 ) = (Fn−1 + Fn )(Fn + Fn+1 ) = Fn+1 Fn+2 . and so the sequence xn := Fn Fn+1 satisfies the modified Fibonacci recurrence xn+1 = xn ⊕ xn−1 . Here we define the first new sequence. Definition 3.4. Let b1 = 0, and for n ≥ 1, b2n = bn b2n+1 ...