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HA7 / Re: HA7-P4
« on: November 05, 2015, 04:49:45 PM »
I think I am also wrong for part d. But general formula is right, just do derivative to get final answer. 
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HA7 / Re: HA7-P7
« on: November 05, 2015, 04:11:45 PM »
Thanks. Hope I get correct answer this time
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HA7 / Re: HA7-P7
« on: November 05, 2015, 03:39:39 PM »
This is my answer for part b. I am not sure whether I do calculation correctly.
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HA7 / Re: HA7-P6
« on: November 05, 2015, 02:07:24 PM »
Solution of part b is attached. Please correct me if I am wrong.
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HA7 / Re: HA7-P4
« on: November 04, 2015, 04:46:40 PM »
Solution for a(i) is attached. Please correct me if I am wrong.
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HA6 / Re: hm6 Q1
« on: October 26, 2015, 02:58:45 PM »
This is my answer for question1(a), the prove part.
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HA4 / Re: Quiz 3 Answers
« on: October 16, 2015, 07:31:26 PM »
Where does 1\3 Ï•(0) come from ?
When I plug numbers into graph which was posted by prof, I do not get this term.
When I plug numbers into graph which was posted by prof, I do not get this term.
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HA3 / Re: HA4-P4
« on: October 04, 2015, 03:15:05 PM »
For (d). The following is my answer.
Given u is continuous when r=0, which means limr→0 u(r,t) exists.
Then limr→0 [f(r+ct) + g(r-ct)] = 0. Because otherwise, limr→0 [f(r+ct) + g(r-ct)] ≠0 implies limr→0 u(r,t) tends to be infinity.
So f(ct) + g(-ct) = 0
f(ct) = - g(-ct)
-f(x) = g(-x)
So g(r-ct) = -f(ct-r). As a result, u = r-1 [f(r+ct) - f(ct-r)]
Given u is continuous when r=0, which means limr→0 u(r,t) exists.
Then limr→0 [f(r+ct) + g(r-ct)] = 0. Because otherwise, limr→0 [f(r+ct) + g(r-ct)] ≠0 implies limr→0 u(r,t) tends to be infinity.
So f(ct) + g(-ct) = 0
f(ct) = - g(-ct)
-f(x) = g(-x)
So g(r-ct) = -f(ct-r). As a result, u = r-1 [f(r+ct) - f(ct-r)]
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HA3 / Re: HA3-P3
« on: October 03, 2015, 09:33:33 PM »
For (b), I am thinking (c1 + c2)2 - 4c1c2 = (c1 - c2)2 which is always larger or equal to zero. And plug in expression(11), it equals to (c1 - c2)2.
Given A2 always larger than 0 (ignore case of equal 0) so B2- 4AC always larger or equal to 0. We dont need to emphasize on this condition.
Therefore, is it possible the answer is A ≠0 ?? I am not sure of this.
Given A2 always larger than 0 (ignore case of equal 0) so B2- 4AC always larger or equal to 0. We dont need to emphasize on this condition.
Therefore, is it possible the answer is A ≠0 ?? I am not sure of this.
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HA3 / Re: HA3-P2
« on: October 03, 2015, 07:41:12 PM »
For (5) I think you miss the last case : x<-ct and x≥ct
In these conditions, I get (-x+ct)\2c
In these conditions, I get (-x+ct)\2c
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