查看完整版本 : Vectors (#1) + Caculus (#2)

Vectors (#1) + Caculus (#2)

[b]Linear Algebra Questions[/b]
My attempt

[[i] 本帖最後由 好男仔 於 2015-9-1 01:17 PM 編輯 [/i]]

[b]Calculus 2 Questions[/b]
Notice that Question 2 is divided into 2 photos
I can't do Question 2b and not sure about 3b as the value should not be negative, I guess.
[b][color=DarkRed]Bear in mind that Calculus 2 is harshly marked on explanations and theory used. So please check my explanations!! Thanks[/color][/b]
My Attempt

[[i] 本帖最後由 好男仔 於 2015-9-1 12:03 PM 編輯 [/i]]

14c14c 2015-9-1 06:57 PM

linear algebra
1你果d唔係linear system of equations黎, 不過是但啦, 你個做法係岩
3d, 佢地唔linearly independent, 唔係basis黎, 最多得3粒架炸

14c14c 2015-9-1 07:23 PM

Calculus
2a你計錯左, 應該係e(ipi/6)(12k+2), 你漏左個i, 所以後面炒晒, e^2pi係唔等於1架
b part應該會用得返a part

Im[2e^((sqrt(3)+i)t/2) /(sqrt(3)+i) ] | pi, 0
= Im[ (sqrt(3)-i)/2  * (ie^sqrt(3)pi/2 - 1) ]
=sqrt(3)e^[sqrt(3)pi/2]/2 + 1/2 (呢個數wolfram左一定無錯)

[[i] 本帖最後由 14c14c 於 2015-9-1 07:38 PM 編輯 [/i]]

14c14c 2015-9-1 07:36 PM

3b a-(a-b) = a-a + b = b

14c14c 2015-9-1 07:42 PM

1.解釋下點解係sqrt(x^2-1)=sinh(theta) 而唔係 -sinh(theta)囉..

[quote]原帖由 [i]14c14c[/i] 於 2015-9-1 06:57 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424827148&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
linear algebra
1你果d唔係linear system of equations黎, 不過是但啦, 你個做法係岩
3d, 佢地唔linearly independent, 唔係basis黎, 最多得3粒架炸 [/quote]
Question 1
What is a linear system of equations? Is it something like below?
x(1,1,1,1)+y(0,2,2,1)+z(1,4,1,1)+w(1,2,1,2)=(0,0,0,0) where x,y,z,w are real numbers?
And then solve for x,y,z,w to see if the system is consistent? If it is, then it is linearly dependent?

Question 3d
What is 3粒? 3 vectors? They are linearly independent, isn't it?
So what should I say? Basis in R^3 only have 3 vectors, so the set is not a basis as it contains 4 vectors.

[quote]原帖由 [i]14c14c[/i] 於 2015-9-1 07:23 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424828898&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
Calculus
2a你計錯左, 應該係e(ipi/6)(12k+2), 你漏左個i, 所以後面炒晒, e^2pi係唔等於1架
b part應該會用得返a part

Im[2e^((sqrt(3)+i)t/2) /(sqrt(3)+i) ] | pi, 0
= Im[  ... [/quote]
Calculus
Question 2a
I got e^(sqrt(3)*t/2)*cos((2pi+3t)/6)

e^2pi係唔等於1 but e^2pi(i)=1 because e^2pi is just a real number, right?

Question 2b
Sorry I dont understand what you are talking about...:smile_27:

[[i] 本帖最後由 好男仔 於 2015-9-2 07:22 PM 編輯 [/i]]

[quote]原帖由 [i]14c14c[/i] 於 2015-9-1 07:36 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424829813&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
3b a-(a-b) = a-a + b = b [/quote]
Question 3b
I dont get what you mean. But I found a careless mistakes. My answer now is 0.5t. Is this correct?

Question 3c
Does my justification sound correct if 3b is correct? I am not confident about this:smile_27:

[[i] 本帖最後由 好男仔 於 2015-9-2 07:19 PM 編輯 [/i]]

[quote]原帖由 [i]14c14c[/i] 於 2015-9-1 07:42 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424830139&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

1.解釋下點解係sqrt(x^2-1)=sinh(theta) 而唔係 -sinh(theta)囉.. [/quote]
Sorry for my poor wording.
Explanation means justify my thinking as well as which method is being used

14c14c 2015-9-3 12:54 AM

linear algebra
1. 係, 個system應該係你寫既咁樣

<=> "x(1,1,1,1)+y(0,2,2,1)+z(1,4,1,1)+w(1,2,1,2)=(0,0,0,0) => x=y=z=w=0"
<=>the system has only one solution
<=> the matrix is at full rank(i.e. rank=4)
3d. a basis by definition is a linearly independent spanning set
{(1,2,3),(2,3,1),(3,1,2)} is a basis, then for any v in R^3
there exist a,b,c in R s.t. v=a(1,2,3)+b(2,3,1)+c(3,1,2)
therefore {(1,2,3),(2,3,1),(3,1,2),v} is not linearly independent, thus not a basis

Calculus
2b.你既第五行
Im[2e^((sqrt(3)+i)t/2) /(sqrt(3)+i) ] evaluate at  t=pi - evaluate at t=0
= Im[ 2/( sqrt(3)+i )  * e^((sqrt(3)+i)t/2) ] evaluate at  t=pi - evaluate at t=0
= Im[ (sqrt(3)-i)/2  * (ie^sqrt(3)pi/2 - 1) ]
= sqrt(3)e^[sqrt(3)pi/2]/2 + 1/2

3a 如果我無睇錯, 應該係
3b at least我睇落去make sense, 但係我對physics既認知係零:smile_13:

[quote]原帖由 [i]14c14c[/i] 於 2015-9-3 12:54 AM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424920079&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]
linear algebra
1. 係, 個system應該係你寫既咁樣

"x(1,1,1,1)+y(0,2,2,1)+z(1,4,1,1)+w(1,2,1,2)=(0,0,0,0) => x=y=z=w=0"
the system has only one solution
the matrix  ... [/quote]
Thanks one more question
how do you go from e^((sqrt(3)+i)pi/2) to ie^sqrt(3)pi/2??
why can you turn the power i into coefficient?

14c14c 2015-9-3 04:08 PM

[quote]原帖由 [i]好男仔[/i] 於 2015-9-3 04:06 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424949147&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

Thanks one more question
how do you go from e^((sqrt(3)+i)pi/2) to ie^sqrt(3)pi/2??
why can you turn the power i into coefficient? [/quote]
e^((sqrt(3)+i)pi/2) = e^(sqrt(3)pi/2  +  ipi/2)=e^(sqrt(3)pi/2)*e(ipi/2)=e^(sqrt(3)pi/2)*i

[quote]原帖由 [i]14c14c[/i] 於 2015-9-3 04:08 PM 發表 [url=http://www.discuss.com.hk/redirect.php?goto=findpost&pid=424949237&ptid=25052362][img]http://www.discuss.com.hk/images/common/back.gif[/img][/url]

e^((sqrt(3)+i)pi/2) = e^(sqrt(3)pi/2  +  ipi/2)=e^(sqrt(3)pi/2)*e(ipi/2)=e^(sqrt(3)pi/2)*i [/quote]
All good:smile_o12:
Cheers man