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Pure questions
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Explain the questions clearly.1. Please explain the bolded part.x/ [ (x - 1)(x - 2)...(x - n) ]= Summation r=1 to n r/ {(r - 1)(r - 2)...[r - (r - 1) ] [ (r + 1) - (r + 2)]...(r - n)(x - r)}2. Let P(x) = 2x^5 + x^3 + 3x^2 + 1 and Q(x) = x^3 + x + 1a) Show that P(x) and Q(x) are relatively prime.b) Find... 顯示更多 Explain the questions clearly. 1. Please explain the bolded part. x/ [ (x - 1)(x - 2)...(x - n) ] = Summation r=1 to n r/ {(r - 1)(r - 2)...[r - (r - 1) ] [ (r + 1) - (r + 2)]...(r - n)(x - r)} 2. Let P(x) = 2x^5 + x^3 + 3x^2 + 1 and Q(x) = x^3 + x + 1 a) Show that P(x) and Q(x) are relatively prime. b) Find the polynomials S(x) and T(x) such that P(x)S(x) + Q(x)T(x) = 1 Using b), resolve 1/ [(2x^5 + x^3 + 3x^2 + 1)(x^3 + x + 1)] into partial fractions. 更新: How come to have [r - (r -1)][(r + 1) - (r + 2)] ? 更新 2: I don't understand why do we need to change [x - (r + 1)] to [(r + 1) - (r + 2)] . 更新 3: No, I don't understand the part ------ [(r + 1) - (r + 2)] . 更新 4: r = Ar(r - 1)(r - 2) ...[r - (r -1)][r - (r + 1)] ...(r - n) -----> r = Ar(r - 1)(r - 2) ...[r - (r -1)][(r + 1) - (r + 2)] ...(r - n), I don't know why! 更新 5: I understand the left part, just don't understand the right part. 更新 6: but the sol has that part. I have to know the reason. As that part is necessary to have the ans -2
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其他解答:
Pure questions
發問:
Explain the questions clearly.1. Please explain the bolded part.x/ [ (x - 1)(x - 2)...(x - n) ]= Summation r=1 to n r/ {(r - 1)(r - 2)...[r - (r - 1) ] [ (r + 1) - (r + 2)]...(r - n)(x - r)}2. Let P(x) = 2x^5 + x^3 + 3x^2 + 1 and Q(x) = x^3 + x + 1a) Show that P(x) and Q(x) are relatively prime.b) Find... 顯示更多 Explain the questions clearly. 1. Please explain the bolded part. x/ [ (x - 1)(x - 2)...(x - n) ] = Summation r=1 to n r/ {(r - 1)(r - 2)...[r - (r - 1) ] [ (r + 1) - (r + 2)]...(r - n)(x - r)} 2. Let P(x) = 2x^5 + x^3 + 3x^2 + 1 and Q(x) = x^3 + x + 1 a) Show that P(x) and Q(x) are relatively prime. b) Find the polynomials S(x) and T(x) such that P(x)S(x) + Q(x)T(x) = 1 Using b), resolve 1/ [(2x^5 + x^3 + 3x^2 + 1)(x^3 + x + 1)] into partial fractions. 更新: How come to have [r - (r -1)][(r + 1) - (r + 2)] ? 更新 2: I don't understand why do we need to change [x - (r + 1)] to [(r + 1) - (r + 2)] . 更新 3: No, I don't understand the part ------ [(r + 1) - (r + 2)] . 更新 4: r = Ar(r - 1)(r - 2) ...[r - (r -1)][r - (r + 1)] ...(r - n) -----> r = Ar(r - 1)(r - 2) ...[r - (r -1)][(r + 1) - (r + 2)] ...(r - n), I don't know why! 更新 5: I understand the left part, just don't understand the right part. 更新 6: but the sol has that part. I have to know the reason. As that part is necessary to have the ans -2
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此文章來自奇摩知識+如有不便請留言告知
(1) x/[(x - 1)(x - 2) ...(x - n)] = ∑[r=1 to n] Ar / (x - r) = ∑[r=1 to n] Ar (x - 1)(x - 2) ...[x - (r -1)][x - (r + 1)] ...(x - n) / [(x - 1)(x - 2) ...(x - n)] The term between the two bolded terms,i.e. (x - r) is not included. Compare denominator, x ≡ ∑[r=1 to n] Ar (x - 1)(x - 2) ...[x - (r -1)][x - (r + 1)] ...(x - n) Put x = r, all term will vanish except for Ar, r = Ar(r - 1)(r - 2) ...[r - (r -1)][r - (r + 1)] ...(r - n) r = Ar(r - 1)(r - 2) ...[r - (r -1)][(r + 1) - (r + 2)] ...(r - n) Ar = r / {(r - 1)(r - 2) ...[r - (r -1)][(r + 1) - (r + 2)] ...(r - n)} (2) P(x) = 2x^5 + x^3 + 3x^2 + 1 Q(x) = x^3 + x + 1 (a) By long division, P(x) = Q(x)(2x^2 - 1) + (x^2 + x + 2) Since the remainder for P(x)/Q(x) is non zero, P(x) and Q(x) are relative prime. (b) Q(x) / (x^2 + x + 2) by long division is (x - 1) with remainder 3 => Q(x) = (x^2 + x + 2)(x - 1) + 3 => (x^2 + x + 2)(x - 1) = Q(x) - 3 Now P(x) = Q(x)(2x^2 - 1) + (x^2 + x + 2) P(x)(x - 1) = Q(x)(2x^2 - 1)(x - 1) + (x^2 + x + 2)(x - 1) P(x)(x - 1) = Q(x)(2x^3 - 2x^2 - x + 1) + Q(x) - 3 P(x)(x - 1) = Q(x)(2x^3 - 2x^2 - x + 2) - 3 P(x)(- x + 1)/3 + Q(x)(2x^3 - 2x^2 - x + 2)/3 = 1 So S(x) = (- x + 1)/3 T(x) = (2x^3 - 2x^2 - x + 2)/3 P(x)S(x) + Q(x)T(x) = 1 1 / [P(x)Q(x)] = [P(x)S(x) + Q(x)T(x)] / [P(x)Q(x)] 1 / [P(x)Q(x)] = S(x)/Q(x) + T(x)/P(x) 1/ [(2x^5 + x^3 + 3x^2 + 1)(x^3 + x + 1)] = [(- x + 1)/3] / (x^3 + x + 1) + [(2x^3 - 2x^2 - x + 2)/3] / (2x^5 + x^3 + 3x^2 + 1) 2009-11-28 11:41:54 補充： Yes, indeed I think there is no need. Simply r - (r + 1) = -1 If you write [r - (r + 1)] I think it is still correct. 2009-11-28 11:52:23 補充： Can you elaborate a bit? I cannot catch your point. What exactly is you question? 2009-11-28 11:56:48 補充： If I have to do the question again without knowing the solution, I will not do this step: r = Ar(r - 1)(r - 2) ...[r - (r -1)][r - (r + 1)] ...(r - n) -----> r = Ar(r - 1)(r - 2) ...[r - (r -1)][(r + 1) - (r + 2)] ...(r - n) The expression on the left hand side is sufficient. 2009-11-28 12:05:43 補充： Ok,, then just ignore the right part. It is not necessary. 2009-11-28 12:20:29 補充： I do not believe you need to care that much. Sometimes the answer can be wrong, unless there are other parts of question that may need to use this. Can you show the full question? Looking at this solution alone, obviously the terms are not balanced. 2009-11-28 12:23:07 補充： (r - 1) (r - 2) .... [r - (r - 1)] [r - (r +1)] ... (r - n) All terms should start with "r -" unless as mentioned above there are other part of the question that needs this special arrangement.其他解答:
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