FP2: De Moivre's theorem and it's applications Watch

username3684524
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So I just finished this chapter (AQA) and my brain is absolutely fried. There is so much to take in and it's so dense with information.

Does anybody have any helpful resources to help with it? I have used Examsolutions which is nice but I'm looking for more.
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RhymeAsylumForever
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https://youtu.be/P2m_u2HVrKs
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diwangislucky
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De Moivre's formula? I thought De Moivre's formula was a direct result of Euler's exponential representation of complex numbers.
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ManLike007
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(Original post by darkforest)
my brain is absolutely fried.
Lmao.

Anyway, building from De Moivre's Theorem.

You learnt the expansions of \cos(n\theta) and \sin(n\theta) in powers of \sin\theta and \cos\theta.

Let's skip ahead and assume you have proved the following,

\cos^{5}\theta = \frac{1}{16}(\cos5\theta + 5\cos3\theta + 10\cos\theta)

Now with Edexcel, there's a chapter called Integration by Reduction (in FP3), this is usually integrating a function involving n, n \in \mathbb{Z^{+}} usually being the power and is high. (IMO Integration by Reduction is faster but it's worth mentioning.)

You can integrate the function \cos^{5}x since you've expanded it,

i.e. \displaystyle \int cos^{5}xdx \Rightarrow \frac{1}{16} \int (\cos5\theta + 5\cos3\theta + 10\cos\theta)d\theta

Alternatively, you'd use Integration by Reduction by setting the power as n, i.e. \displaystyle \int \cos^{5}xdx \Rightarrow I_{n} = \int \cos^{n}x dx = \int \cos^{1}x \cdot \cos^{n-1}dx then use integration by parts etc. (I won't get into it as this isn't related to complex numbers but it's the idea of integrating a function to a high power as shown).

Pretty much the only new thing I've learnt after A Level regarding De Moivre's Theorem. The rest is Euler's exponential representation.
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DFranklin
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(Original post by ManLike007)
(IMO Integration by Reduction is faster but it's worth mentioning.)
I know this is taking things even more off topic, but FWIW, to integrate \cos^n x where n is odd, the fastest method is to rewrite as:

\cos x (1-\sin^2 x)^{(n-1)/2} and then integrate by recognition.

e.g. \int \cos^5 x \,dx = \int \cos x (1-\sin^2 x)^2
=\int \cos x (1 - 2 \sin^2 x + \sin^4 x) \,dx =  \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C
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DFranklin
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(Original post by darkforest)
So I just finished this chapter (AQA) and my brain is absolutely fried. There is so much to take in and it's so dense with information.

Does anybody have any helpful resources to help with it? I have used Examsolutions which is nice but I'm looking for more.
I'm not sure what you really need to remember for this other than

(\cos x + i \sin x)^k = \cos kx + i \sin kx)

and the two main applications are:

you can expand (\cos + i \sin x)^k binomially to get expressions for cos kx and sin kx in terms of cos x and sin x.

Conversely, setting z = cos x + i \sin x, then z+(1/z) = 2 cos x and z-1/z = 2i sin x; you can use this and expand (z+1/z)^k (or (z-1/z)^k binomially to get expressions for cos^k x and sin^k x as a linear combination of basic trig expressions (e.g. cos^3 x = (cos 3x + cos x)/4), which is handy for reducing to something that's easy to integrate.

I'm not sure what else you think you need to know?
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ManLike007
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(Original post by DFranklin)
I know this is taking things even more off topic, but FWIW, to integrate \cos^n x where n is odd, the fastest method is to rewrite as:

\cos x (1-\sin^2 x)^{(n-1)/2} and then integrate by recognition.

e.g. \int \cos^5 x \,dx = \int \cos x (1-\sin^2 x)^2
=\int \cos x (1 - 2 \sin^2 x + \sin^4 x) \,dx =  \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C
I'm assuming this also works for \sin{x} for odd values of n. It's interesting but I believe there is a doable limit you'd use this 'trick' as a power of \frac{n-1}{2} means any number, I mean if you had \displaystyle \int \cos^{9}x \,dx \Rightarrow \int \cos x (1 -sin^{2}x)^{4} \,dx, you know, it's not really fun to expand now is it? (Unless you can do Binomial expansion in your head then fair play to you sir)

(Apologies to OP for swaying this off even more)
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username3684524
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Thanks for all of the replies!

It seems like I was really just struggling with the concept of the ‘n nth roots of complex numbers’.

I did a bit more work on it and I seem to get it now.
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DFranklin
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(Original post by ManLike007)
I'm assuming this also works for \sin{x} for odd values of n. It's interesting but I believe there is a doable limit you'd use this 'trick' as a power of \frac{n-1}{2} means any number, I mean if you had \displaystyle \int \cos^{9}x \,dx \Rightarrow \int \cos x (1 -sin^{2}x)^{4} \,dx, you know, it's not really fun to expand now is it? (Unless you can do Binomial expansion in your head then fair play to you sir)
Well, yes, I can (and you should be able to for a number this small).

So integral is: S - 4S^3/3 +6S^5/5 -4S^7/7 + S^9 / 9, where S = sin x.

Which I literally wrote down.

Note that using a reduction formula is also not really fun when n is large (except for certain special cases where the limits mean it works out unusually nicely).

There are still times where you want to use a reduction anyhow (particularly when you want to examine the behaviour as n->infinity), but for a standard question, it saves a bit of time.
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username3555092
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(Original post by DFranklin)
I know this is taking things even more off topic, but FWIW, to integrate \cos^n x where n is odd, the fastest method is to rewrite as:

\cos x (1-\sin^2 x)^{(n-1)/2} and then integrate by recognition.

e.g. \int \cos^5 x \,dx = \int \cos x (1-\sin^2 x)^2
=\int \cos x (1 - 2 \sin^2 x + \sin^4 x) \,dx =  \sin x - \frac{2}{3} \sin^3 x + \frac{1}{5} \sin^5 x + C
PRSOM for the nifty method, but did you really think that A Level maths exams would afford you opportunities to use whichever method you wished? If that were the case I'd probably have used modular arithmetic instead of induction for all the divisibility questions in FP1.

Attached to my post is how a similar integral would appear on an exam paper.
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Sir Cumference
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(Original post by I hate maths)
PRSOM for the nifty method, but did you really think that A Level maths exams would afford you opportunities to use whichever method you wished? If that were the case I'd probably have used modular arithmetic instead of induction for all the divisibility questions in FP1.

Attached to my post is how a similar integral would appear on an exam paper.
Well once part a) has already been done, the quickest way is to use it instead of using DFranklin's method
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username3555092
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(Original post by Notnek)
Well once part a) has already been done, the quickest way is to use it instead of using DFranklin's method
Which is true, but my point being that they often force part a on you.
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DFranklin
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(Original post by I hate maths)
Attached to my post is how a similar integral would appear on an exam paper.
I think you need to change "would" with "might".

But yeah, if they tell you to use a stupid method, you have to use a stupid method. (And I don't think there's much doubt that there are at least 2 methods *significantly* quicker than the one they tell you to use).
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Sir Cumference
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(Original post by DFranklin)
But yeah, if they tell you to use a stupid method, you have to use a stupid method. (And I don't think there's much doubt that there are at least 2 methods *significantly* quicker than the one they tell you to use).
While this is true for a lot of A Level questions, I think for the one posted above it's really testing that a student can apply De Moivre's theorem. The integration is just a "nice" (in their opinion) follow up and isn't really testing that much integration ability.
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DFranklin
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(Original post by Notnek)
While this is true for a lot of A Level questions, I think for the one posted above it's really testing that a student can apply De Moivre's theorem. The integration is just a "nice" (in their opinion) follow up and isn't really testing that much integration ability.
Yes sure. Given how the question is broken up, it's the natural way to do it. But if you look at it as "solve this integral, here's a hint for the first step" (i.e. if you assume the goal was to find the integral), then it's a reasonably inefficient way of going about it.

[More generally, I think the (z+1/z)^k trick is used very rarely outside of this kind of A-level question, so I find it frustrating that they try to pretend it's actually "equivalently useful" to the (c+is)^k trick. I'd guess I've used (c+is)^k ten times more often that the (z+1/z) trick with real problems.]
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