What are the equivalents of (A' n B') , (A n B)U (A n B') , and [A' n (A u B)]' ? Please help me. Thanks.
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(A'nB')=(AUB)'\n(AnB)U(AnB')=A\n[A'n(AuB)]'=AUB'=(B&#92;A)'\n\n(B&#92;A) means B which is not contained in A; B-(AnB)
1109004
In general any lanuages with words of unbounded length ,where the same pattern is repeated in form or length, usually are not regular.\n\nI'm sorry I'm not familiar with Myhil Nerode, so if an answer depends upon that, I can't help you.\n\nL1 is not regular. Proven via the Pumping Lemma.\n\nSince wcw is unbounded, if it were a regular expression, it would need to accept regular expressions of the form uv*w. NOTE: The w here is not the same as the w in wcw. I used uv*w, because that's the tradition in the Pumping Lemma. Let me replace uv*w with UV*W to keep the two straight.\n\nSince wcw needs to accept UV*W, where can the "c" reside in UV*W?\n\nCan it be in U? No, because then the V would pump a bunch of a's and b's to place accept string with more characters on the right of w than the left.\n\nCan w be in V? No, because pumping the w within V would accept more than on w (or none.)\n\nCan w be in W? No for the same reason it can't be in U.\n\nL2 is regular. It's just another way of stating that you have an even number of a's and b's. The don't have to be the same, and I don't think you need to know where the xy split is. You could "implement" an FA with 2 states. One state is the start state, and the other is the final state. The transitions for a and b just hop back and forth between the two. I think that's all you need to show.\n\nL3 is not regular using the pumping lemma.\n\nSince a^n (where n is prime) is unbounded, it needs to accept a set of a's in the form of uv*w. Consider what happens you when pump v the following number of times |u| + |w|.\n\nYou'll get a string of a's of the following length:\n\n|u| + |v| * (|u| + |w|) + |w|\n\nThe first |u| comes from the u. The last |w|  comes from the w. The middle |u| + |w| comes from the pump of v which is of length |v|. To make it easier, let's use capital letters for length.\n\nTherefore the length of the string accepted is:\n\nU + V(U+W) + W\n\nU + V*U + V*W + W\n\nU(1 + V) + W(1 + V)\n\n(U + W) * (1 + V)\n\nAha. The length of the string is a produce of two integers. Therefore it is not prime. We've pumped uv*w to a string that can't be prime.\n\nL4 - No idea. I suspect this may be where Myhil Nerode comes in.
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the 1st three terms means n=1, n=2, n=3 so substitute\n\nThis gives, 3*1+2,3*2+2,3*3+2\n\n=5,8,11
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1)1,8,27,64,125\n  x^3 = 1^3, 2^3, 3^3, 4^3, 5^3\n\n2)9, 13, 15, 19, 21, 25\n a(n)+ n-2 = {a(n-1) +a(n+1)}/2\n\n3)00001, 00011, 00101, 00111, 01001\n  1,3,5,7,9 in binary\n\n4)17,23,5,18,20,25,21,9,15,16\nok, your teacher is an ass for this one..  heres the breakdown\nq is the 17th letter, w is the 23 letter, e, r, t, y, u, i, o, p\nthe top row of a standard U.S. keyboard. i think i earned 10 points for that one alone\n\n\n5)\nFOUR\nF I V E\nN I N E\n\nR=0\nF=2\nI=3\nN=5\nE=6\nU=7\nV=8\nO=9\n\n2970 + 2386 = 5356\nDONE.  Hope this works for you
998387
half will be A\nhalf will be B\nnon will be o\nnone will be ab
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2 + (a + b)\ngiven\n\n2 + (b + a)\ncommutative prop. of addition\n\n(2 + b) + a\nassociative prop. of addition\n\n(b + 2) + a\ncommutative prop. of addition\n\nb + (2 + a)\nassociative prop. of addition\n\n^_^
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<> means 'not equal to'\n\nThe C/C++ equivalent is != \n\n a <> b : means 'a is not equal to b'\n 5 <> 6 : gives the value True\n 5 <> 5 : gives the value False
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1= B\n2= A\n3= D\n4= D\n5= D\n6= C
904637
n + 8in. < 24in.\n    - 8in.    - 8in.\n          n< 16in.
819996
T1=m1*a, no matter are there m2 & m3 or not.\nT2=(m1+m2)*a, no matter whether m1 & m2 are connected with a string or are in one lump, the presence of m3 being unimportant too.
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Without knowing the relationship between successive elements, I think the best you can do is this:\n[k=0, n-1]∑(x_(k+1) - x_k)²\n[k=0, n-1]∑(x_(k+1)² - 2x_k*x_(k+1) + x_k²)\n[k=0, n-1]∑x_(k+1)² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n[k=1, n]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1)) + [k=0, n-1]∑x_k²\n\nx_0 + x_n + 2[k=1, n-1]∑x_k² - 2[k=0, n-1]∑(x_k*x_(k+1))
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Since B is in the middle, lets set B as 0 for example purposes.  A would be any negative number (A<B) and C would be any positive number (B<C).  Since there are no 'or equal to' symbols, A<C.  This is true for all possilbe cases.
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If by a(i) and b(i) you mean "a times i" and "b times i", then:\n\nFor i = 0:\n\na*i / (b*i + 1) = 0 / 1 = 0\n\nSo you have:\n\nPo = 1 / (1 + 0 + (0 * (a / (b + 1))) + (0 * (a / (b + 1)) * (2a / (2b + 1))) + ...)\n\nPo = 1 / 1\n\nPo = 1\n\nIf by a(i) and b(i) you mean a and b are infinite sequences (then presumably a / b means "for all i, a(i) / b(1) < 1"), well then that is something else entirely.\n\nThis reminds me of the problem to expand and simplify the product of twenty-six binomial terms:\n\n(x - a) (x - b) (x - c) ... (x - z)
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LOL\n\nI can follow you up to n+1=2, what's 2n?\n\nn+1=2  so  n=2-1  so  n=1  so  2n=2
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Sorry I'm not english spoken\n\nIl suffit de choisir deux vecteurs u et v non nuls et non liés.\nu ∧ v ≠ 0\nOn constate alors que :\n(u ∧ u) ∧ v = 0 ∧ v = 0\nSupposons que :\nu ∧ (u ∧ v) = 0\nAlors\nu et u ∧ v sont liés\nCe qui est absurde car :\nu ∧ v ⊥ u\nDonc :\nu ∧ (u ∧ v) ≠ 0\nD'où :\n(u ∧ u) ∧ v ≠ u ∧ (u ∧ v)
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G pairs with C and A pairs with T, so its GAATCCGAATGGTA\n\nI assume the "h" at the end is a typo?\n__________________________________________\nsouthernserendipiti wrote:\n>>>What's the "th" at the end?\nI assumed it was a t and a typo : )
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I do not believe this is always true, only in the case of commuting matrices.\n\nLet A'=A\nLet B'=B\n\nThen (AB)' = B'A' = BA\n\nBut AB is  not always equal to BA unless A and B commute.