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Further Mathematics T Paper 1 讨论专区
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发表于 22-6-2011 05:59 PM
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请问怎样这个怎么做? simplify [2-√3i]^6 using de Moivre's theorem.
Wangs 发表于 20-6-2011 05:42 PM 
我们必须把它换成 r[cosx + i sinx] 的form 是吗?可是我拿到的x 是 很多decimal place 的,所以不知道怎么继续。 |
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发表于 22-6-2011 06:02 PM
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发表于 22-6-2011 06:37 PM
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发表于 22-6-2011 07:46 PM
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回复 81# Wangs
方便给我你的facebook email吗? |
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发表于 22-6-2011 10:10 PM
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发表于 22-6-2011 10:23 PM
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回复 83# yuanhua88
其实我还是不是很明白,请问你可以给我一些具体的例子吗? 因为我如果没有算cos nx 和sin nx, 只算cos x 和 sin x 也拿不到答案。 |
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发表于 25-6-2011 01:17 AM
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回复 86# Wangs
从我分享的link那里可以看到,cos nx 其实是可以通过cos x 来计算
打个例子,[2-√3i]^6, cos x=2/sqrt(7)

cos 6x= 32*(2/sqrt(7))^6-48*(2/sqrt(7))^4+18(2/sqrt(7))^2-1
=-0.42 |
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发表于 30-6-2011 03:25 AM
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本帖最后由 50912cmea 于 3-3-2012 12:44 PM 编辑
8. Crawshaw, J. & Chambers, J., A concise Course in Advanced Level Statistics (Fourth Edition), Nelson Thornes, 2001
http://www.mediafire.com/?m35cnaqq5nbjdxs
2. Gaulter, B. & Gaulter, M., Further Pure Mathematics, Oxford, 2001
http://www.mediafire.com/?pf5xxssp3frehiv
6. Rosen, K. H., Discrete Mathematics and Its Applications (Fifth Edition), McGraw-Hill, 2003
http://www.mediafire.com/?2pxxqqnozzfuwb1
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发表于 12-7-2011 08:17 PM
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请问日新和槟城钟灵的老师在考试时会让拿further maths 的学生考试吗?我指的是校内的。 |
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发表于 4-9-2011 10:41 AM
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回复 88# 50912cmea
我觉得旧版FMT课本的stats部分,好过A concise course in A-level stats.... |
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发表于 4-9-2011 04:16 PM
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发表于 5-9-2011 02:31 AM
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回复 50912cmea
我觉得旧版FMT课本的stats部分,好过A concise course in A-level stats....
cheesit92 发表于 4-9-2011 10:41 AM 
这个 A concise course in A-level stats 是 4th edition, 之前 2nd edition 比较多东西。。。
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发表于 7-9-2011 10:02 PM
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发表于 25-9-2011 08:03 AM
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发表于 22-10-2011 06:31 PM
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请问 graph theory 的 cycle 需要写vertices 吗? example:v1, e1, v2, e2
还是只是写edges就可以了。 |
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发表于 2-1-2012 07:40 PM
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发表于 2-1-2012 07:42 PM
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Also, anyone who got the past year from hebe, why not upload it and post the links here? That should help prevent those material from being 'lost'. |
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发表于 14-1-2012 10:03 PM
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请问怎样 integrate (arcsinh x)^2 ? |
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发表于 22-1-2012 11:09 PM
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这个问题够复杂的
先是substitution
然后还要两次integration by parts
Possible intermediate steps:
integral sinh^(-1)(x)^2 dx
For the integrand sinh^(-1)(x)^2, substitute u = sinh^(-1)(x) and du = 1/sqrt(x^2+1) dx:
= integral u^2 cosh(u) du
For the integrand u^2 cosh(u), integrate by parts, integral f dg = f g- integral g df, where
f = u^2, dg = cosh(u) du,
df = 2 u du, g = sinh(u):
= u^2 sinh(u)-2 integral u sinh(u) du
For the integrand u sinh(u), integrate by parts, integral f dg = f g- integral g df, where
f = u, dg = sinh(u) du,
df = du, g = cosh(u):
= u^2 sinh(u)-2 u cosh(u)+2 integral cosh(u) du
The integral of cosh(u) is sinh(u):
= u^2 sinh(u)+2 sinh(u)-2 u cosh(u)+constant
Substitute back for u = sinh^(-1)(x):
= -2 sqrt(x^2+1) sinh^(-1)(x)+2 x+x sinh^(-1)(x)^2+constant |
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发表于 23-1-2012 09:18 PM
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谢谢你 |
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